最新《二次根式》培优试题及答案

发布时间:2019-12-01 11:05:32

《二次根式》提高测试

(一)判断题:(每小题1分,共5分)

143356dd3e1f1050c2531804b5ea4ba3f.png=-29b60fb9f15ff81cd3f96ee290618864d.png…………………(  )【提示】8ffb42436708d28fc47f8b2b0f3c0594.png|2|2.【答案】×.

291a24814efa2661939c57367281c819c.png2的倒数是91a24814efa2661939c57367281c819c.png2.(  )【提示】c155529df06092d4e25ba8393d2dbc04.png42bd69154912310deef6f3cbcbaf1c00.png=-(91a24814efa2661939c57367281c819c.png2).【答案】×.

3eafac7bcb7202f70f2086863e4c180d5.png10013d5528fe91d640572e17331aa6f2.png(  )【提示】eafac7bcb7202f70f2086863e4c180d5.png|x1|10013d5528fe91d640572e17331aa6f2.pngx1x1).两式相等,必须x1.但等式左边x可取任何数.【答案】×.

49b60fb9f15ff81cd3f96ee290618864d.png7964c6a339acf2ddea25a5ef0552b97e.png775885c53de53f10618b609f87131b45.pngf1148fe2cd67063c1e9e0687a53ef53e.png是同类二次根式.(  )【提示】7964c6a339acf2ddea25a5ef0552b97e.png775885c53de53f10618b609f87131b45.pngf1148fe2cd67063c1e9e0687a53ef53e.png化成最简二次根式后再判断.【答案】√.

5a4fcf20a5d47c68d6e64ba33f53d3e52.png7e9892c37854c2c741ac0877f6ed0148.png3af8e0e551b774e3e608d7acdc7f9802.png都不是最简二次根式.(  )3af8e0e551b774e3e608d7acdc7f9802.png是最简二次根式.【答案】×.

(二)填空题:(每小题2分,共20分)

6.当x__________时,式子bddc4513800156dd547150dfba63647f.png有意义.【提示】1a03d0f7242823c05e0f16ad19f85201.png何时有意义?x0.分式何时有意义?分母不等于零.【答案】x0x9

7.化简-f78a57ca094de2afc2c3413fb815f08b.png875f301244667a1a8fdab97fb26d6c66.png÷9f777348b96f1c9037c8e224c16774b2.png_.【答案】-2ac17e049691f2eb2d93102b7d4b309572.png.【点评】注意除法法则和积的算术平方根性质的运用.

8a638ed7bdb77a19d73ec0973f0aa9c73a.png的有理化因式是____________.【提示】(a638ed7bdb77a19d73ec0973f0aa9c73a.png)(________)=a29190da5b4d145457c87049546f66dc0c.pnga638ed7bdb77a19d73ec0973f0aa9c73a.png.【答案】a638ed7bdb77a19d73ec0973f0aa9c73a.png

9.当1x4时,|x4|cde3065f3ea5ba24818b7c6d34925f75.png________________

【提示】x22x1=(  )2x1.当1x4时,x4x1是正数还是负数?

x4是负数,x1是正数.【答案】3

10.方程d21848cdd835abcb491be1f151e9b6c6.pngx1)=x1的解是____________.【提示】把方程整理成axb的形式后,ab分别是多少?c8168f582ccafb1cc49f0ef6d5a66f3b.png4bc3c43b38c136cb7a40b233c50b591e.png.【答案】x32d21848cdd835abcb491be1f151e9b6c6.png

11.已知abc为正数,d为负数,化简84b5a358f9cd74ba7cda43fc9f4fb90c.png______.【提示】9ce233c5cc6a28403981a452b0ec11c0.png|cd|=-cd

【答案】9b60fb9f15ff81cd3f96ee290618864d.pngcd.【点评】∵ ab6da5183d6a9bdf02efafe5c56bc9391e.pngab0),∴ abc2d2=(cb6290652142cb5d190d9414cbdc5845.png)(f3a8fc058fa60df5c476a6dd77de0bdb.png).

12.比较大小:-3a4f27e71acfd642035c19445404269d.png_________55126ef0c267c33b415e34ddd3878431.png.【提示】21801cfc88edd59ca7296ac197514e703.pngb5e6268f8a3db07e7c23ff5d9bb1c516.png491a24814efa2661939c57367281c819c.png9cbb0911d1ed1ce601b56b66aefbf7ed.png

【答案】<.【点评】先比较b5e6268f8a3db07e7c23ff5d9bb1c516.png9cbb0911d1ed1ce601b56b66aefbf7ed.png的大小,再比较77335f073853c71264497487f426486e.png15964d242cdc64df329e9baac0abc1b6.png的大小,最后比较-77335f073853c71264497487f426486e.png与-15964d242cdc64df329e9baac0abc1b6.png的大小.

13.化简:(75d21848cdd835abcb491be1f151e9b6c6.png)2000·(75d21848cdd835abcb491be1f151e9b6c6.png)2001______________

【提示】(75d21848cdd835abcb491be1f151e9b6c6.png)2001(75d21848cdd835abcb491be1f151e9b6c6.png)2000·_________[75d21848cdd835abcb491be1f151e9b6c6.png]

75d21848cdd835abcb491be1f151e9b6c6.png·(-75d21848cdd835abcb491be1f151e9b6c6.png)=?[1]【答案】-75d21848cdd835abcb491be1f151e9b6c6.png

【点评】注意在化简过程中运用幂的运算法则和平方差公式.

14.若3d0f16f253ed7418df5cea1d414dce0b.pngacde70f9e0a45561eb92636defc65652.png0,则(x1)2(y3)2____________.【答案】40

【点评】3d0f16f253ed7418df5cea1d414dce0b.png0acde70f9e0a45561eb92636defc65652.png0.当3d0f16f253ed7418df5cea1d414dce0b.pngacde70f9e0a45561eb92636defc65652.png0时,x10y30

15xy分别为849d66b49a2741e5b36e82bc1a9d6a14b.png的整数部分和小数部分,则2xyy2____________

【提示】∵ 349d66b49a2741e5b36e82bc1a9d6a14b.png4,∴ _______849d66b49a2741e5b36e82bc1a9d6a14b.png__________[45].由于849d66b49a2741e5b36e82bc1a9d6a14b.png介于45之间,则其整数部分x=?小数部分y=?[x4y449d66b49a2741e5b36e82bc1a9d6a14b.png]【答案】5

【点评】求二次根式的整数部分和小数部分时,先要对无理数进行估算.在明确了二次根式的取值范围后,其整数部分和小数部分就不难确定了.

(三)选择题:(每小题3分,共15分)

16.已知4caa2c6687c24f1245b461a0dad13f6d.png=-xa365e380ae9ea9a71431ea0081d25246.png,则………………(  )

Ax0   (Bx≤-3   (Cx≥-3   (D)-3x0【答案】D

【点评】本题考查积的算术平方根性质成立的条件,(A)、(C)不正确是因为只考虑了其中一个算术平方根的意义.

17.若xy0,则bf7f1deff897878b344190b71e4ab9c1.png0a1fce48fb8b53a148ffa7826cf66ddc.png=………………………(  )

A2x   B2y   C)-2x   D)-2y

【提示】∵ xy0,∴ xy0xy0

∴ bf7f1deff897878b344190b71e4ab9c1.pngee3e541a0e03424cfccdfecd0de1a76f.png|xy|yx

0a1fce48fb8b53a148ffa7826cf66ddc.png8b655c7a7f1b8b32e66976eb4761717d.png|xy|=-xy.【答案】C

【点评】本题考查二次根式的性质c18ecf8c296ffe653d0909ec8574a843.png|a|

18.若0x1,则788fc2ad72cda0f63ba624eeb04d212d.png2ba7904d822656ebd865093c55825b2c.png等于………………………(  )

Abcb5585525348a4cf25a8713cf427026.png   (B)-bcb5585525348a4cf25a8713cf427026.png   (C)-2x   D2x

【提示】(xafc48b56873694f3d43097841ecc3f4f.png)24(xafc48b56873694f3d43097841ecc3f4f.png)2(xafc48b56873694f3d43097841ecc3f4f.png)24(xafc48b56873694f3d43097841ecc3f4f.png)2.又∵ 0x1

∴ xafc48b56873694f3d43097841ecc3f4f.png0xafc48b56873694f3d43097841ecc3f4f.png0.【答案】D

【点评】本题考查完全平方公式和二次根式的性质.(A)不正确是因为用性质时没有注意当0x1时,xafc48b56873694f3d43097841ecc3f4f.png0

19.化简7cd76b24f4dd8d26047d708e2bc7c7be.png84c40473414caf2ed4a7b1283e48bbf4.pnga09371d7a2e3ae86a00aab4771e39d255d.png………………………………………………………………(  )

A82d43618f1d073cc3dc6b92b411f9194.png   (B)-c17e049691f2eb2d93102b7d4b309572.png   (C)-82d43618f1d073cc3dc6b92b411f9194.png   (Dc17e049691f2eb2d93102b7d4b309572.png

【提示】d50acc03581b84b815cee3ad723008e4.pngb37f8fa9c8a2aa82b9007dc4de792f0a.png82d43618f1d073cc3dc6b92b411f9194.png·c18ecf8c296ffe653d0909ec8574a843.png|a|82d43618f1d073cc3dc6b92b411f9194.png=-a82d43618f1d073cc3dc6b92b411f9194.png.【答案】C

20.当a0b0时,-a29b60fb9f15ff81cd3f96ee290618864d.pngb可变形为………………………………………(  )

Aac0b641b58c72f34e96e7b86b412f4a3.png (B)-fc700fb63299233d63eff2709262dc86.png (C0490490f15e3acff9fe820624ded2f22.png (Dbfa6515f5e47ab6cc79e4cd374caa23c.png

【提示】∵ a0b0

∴ -a0,-b0.并且-a16ce9e725a0ef77882e77f08d353ca28.png,-b43c54bc2f5cc079051c87c4ae8d70605.png9b60fb9f15ff81cd3f96ee290618864d.png32a990dfb68d0dcdb885fd8f2fe05003.png

【答案】C.【点评】本题考查逆向运用公式252c57d1f8e3d8aed72e8b1c262290ef.pngaa0)和完全平方公式.注意(A)、(B)不正确是因为a0b0时,c17e049691f2eb2d93102b7d4b309572.pnge35df863fc94d6c6fa5b34788e170373.png都没有意义.

(四)在实数范围内因式分解:(每小题3分,共6分)

219x25y2;【提示】用平方差公式分解,并注意到5y2f03fc3c7f8eb497b8d51ceff301dff1c.png.【答案】(3xaa4e3cfb024c7ff30a8846913966dfb1.pngy)(3xaa4e3cfb024c7ff30a8846913966dfb1.pngy).

224x44x21.【提示】先用完全平方公式,再用平方差公式分解.【答案】(d21848cdd835abcb491be1f151e9b6c6.pngx1)2(d21848cdd835abcb491be1f151e9b6c6.pngx1)2

(五)计算题:(每小题6分,共24分)

23.(2efd0c951ff8242b0ef80a18731c6e48.png)(1b09e9cabad5c464df653d62189ab768.png);

【提示】将5116381b490edb6a4a1c7e3af5654432.png看成一个整体,先用平方差公式,再用完全平方公式.

【解】原式=(5116381b490edb6a4a1c7e3af5654432.png)2f2198cb9b137abf2cba6aa369620c1ca.png524a68ac0136e5c8446ca046abe9e88800.png32624a68ac0136e5c8446ca046abe9e88800.png

240cb3bcdf486654dca8ad8efd43e52f03.png4102ab684dc955240b272326c71f0563.png87f3d5f23b5be887252ca598cf97d8c7.png;【提示】先分别分母有理化,再合并同类二次根式.

【解】原式=da5bf8ce29ed321d8ac1870ff3d0bdda.png828a9d0fa391ff7a6a3cc7481c217791.png6531c1a3c2528cfe9a9cc041bbf3bbe4.png449d66b49a2741e5b36e82bc1a9d6a14b.png49d66b49a2741e5b36e82bc1a9d6a14b.png1801cfc88edd59ca7296ac197514e703.png31801cfc88edd59ca7296ac197514e703.png1

25.(a26dabdcf8d8716f589654771d19c95138.png17758d38625d268176921641178cf417.png9f0d528e07bddb8598dcd2443e67b6c3.pngd049603d2ecab82f77394e4fa3814b12.png5c257326242086b47ccab8956da94b2d.png)÷a2b26dabdcf8d8716f589654771d19c95138.png

【提示】先将除法转化为乘法,再用乘法分配律展开,最后合并同类二次根式.

【解】原式=(a26dabdcf8d8716f589654771d19c95138.png17758d38625d268176921641178cf417.png9f0d528e07bddb8598dcd2443e67b6c3.pngd049603d2ecab82f77394e4fa3814b12.png5c257326242086b47ccab8956da94b2d.png·738263f66d36a8870bf8a87be30616ac.png5c257326242086b47ccab8956da94b2d.png

06ec3a34b58f5b919636e281c008a2bd.png8417792d2333414ef5bf3269ed6ea574.pngc0a2649c2900afb4d94d77ce59eec882.pngf0bf0de6aec414eab2bc23c5a26023c0.png5c677dde95faecf0def3977c74047e83.pngfe703b84a0277a8a4323fef487d96e33.png

06ec3a34b58f5b919636e281c008a2bd.pnge235f2511c3aa74329cd97e549ad7b91.png738263f66d36a8870bf8a87be30616ac.pngabfa869e16886e5c071a2d95f810178e.png

26.(c17e049691f2eb2d93102b7d4b309572.pngf674fe39a8682a0f5b42edf09a90e177.png)÷(b1897a4d2b24edfa09d1e6a330fb2c92.png5b3dc27c6e005667bf72796249ed45d4.png22fe98d0987f4c0e9917cf49d515f783.png)(ab).

【提示】本题应先将两个括号内的分式分别通分,然后分解因式并约分.

【解】原式=7122a85436d65a97cfa424b2a6627bb0.png÷1ab36212bbfcb28a5bf830217d442368.png

f9e9f7fe816229f8893c20f28c48c66b.png÷8ef4e8bcff119cd24cc22925b8c57a71.png

f9e9f7fe816229f8893c20f28c48c66b.png·981256c7522b36ece0958032aa127e31.png=-61457c606a4e9a373cb95b6819085e6f.png

【点评】本题如果先分母有理化,那么计算较烦琐.

(六)求值:(每小题7分,共14分)

27.已知x2263f20e10353213cc15eb464c6d1720.pngy351bbfc9b1f2aad1a51c2675200dff1a.png,求e7a1e73f00446961e1542b7d03befe0c.png的值.

【提示】先将已知条件化简,再将分式化简最后将已知条件代入求值.

【解】∵ x2263f20e10353213cc15eb464c6d1720.png445c9bd55391b81160da9b3c3fa1a222.png5265ebe73c520528b6825b8ff4002086d7.png

y351bbfc9b1f2aad1a51c2675200dff1a.png76691cc2a609a58301516ea50bbcbc02.png5265ebe73c520528b6825b8ff4002086d7.png

∴ xy10xy465ebe73c520528b6825b8ff4002086d7.pngxy52(265ebe73c520528b6825b8ff4002086d7.png)21

e7a1e73f00446961e1542b7d03befe0c.pngaab47470ace65d95f753735faf410667.pnga5fd12ee88557c7d71d8eb1b5b57de2c.pnge397aa185c53837cf9054e32c487f9fd.png8352afa89545386466a2cd92c3d65054.png

【点评】本题将xy化简后,根据解题的需要,先分别求出“xy”、“xy”、“xy”.从而使求值的过程更简捷.

28.当x1d21848cdd835abcb491be1f151e9b6c6.png时,求f7300b37713782d95ad9a5b9691fd28d.png32fd3348a1e37ff0c87c86c6769f45bf.png58aa21f036f87b0352c42fd5e8f0f529.png的值.

【提示】注意:x2a2a1388434a5fe6a235c5f755ea0ebf1e9.png

∴ x2a2xa5a6d1ea584db44e6dfd9b28e74aeea9.pnga5a6d1ea584db44e6dfd9b28e74aeea9.pnga5a6d1ea584db44e6dfd9b28e74aeea9.pngx),x2xa5a6d1ea584db44e6dfd9b28e74aeea9.png=-xa5a6d1ea584db44e6dfd9b28e74aeea9.pngx).

【解】原式=f3d1e6103cb67b88f67dfc8d3e0a3bfe.pngb38643ce953ad7848eb14c95e7dcd841.png58aa21f036f87b0352c42fd5e8f0f529.png

e79ad0ece04e59cc9afc13c2d2e7d17b.png

0c8d524b026144f4e81c348d529b8098.png=4812f334dcb891a12b20289e12fb5842.pngc2975d0a032dfa9b8d29ee4901a1c5a2.png

afc48b56873694f3d43097841ecc3f4f.png.当x1d21848cdd835abcb491be1f151e9b6c6.png时,原式=bfd6d855151d9c2c8f187ebc9a9ff311.png=-1d21848cdd835abcb491be1f151e9b6c6.png.【点评】本题如果将前两个“分式”分拆成两个“分式”之差,那么化简会更简便.即原式=f3d1e6103cb67b88f67dfc8d3e0a3bfe.pngb38643ce953ad7848eb14c95e7dcd841.png58aa21f036f87b0352c42fd5e8f0f529.png

3fb8dbe5ac3b3d7ceb31ae6113ed9740.pngada1ddaa919d610b49b4b3a59b16dc7e.png58aa21f036f87b0352c42fd5e8f0f529.pngafc48b56873694f3d43097841ecc3f4f.png

七、解答题:(每小题8分,共16分)

29.计算(2aa4e3cfb024c7ff30a8846913966dfb1.png1)(5bb91d379fc17772ce797598f9d5491c.pnge8077b3dd7246070d199e5a34acb0a97.pngba5a128edeb37ed29b99fb4f5eeca603.png+…+174f6ffe0a8045b42ee17eb6216e2408.png).

【提示】先将每个部分分母有理化后,再计算.

2、传统文化对大学生饰品消费的影响【解】原式=(2aa4e3cfb024c7ff30a8846913966dfb1.png1)(ab05a7a77efecfd60e8723b91c72a42b.png13f7c6a032591936ae97f12627d37517.pngd673a21a835f4379bc153fc2ef2508c2.png+…+ad2464646b0329d4debc1cac817e2c2f.png

=(2aa4e3cfb024c7ff30a8846913966dfb1.png1[c8168f582ccafb1cc49f0ef6d5a66f3b.png)+(ca34deae0b82b9d84af6fd79a2f4714b.png)+(baf9a46ffe0be9e2a272686c9f581c88.png)+…+(47b561ac96b78ad0d5a2cfcc92d793aa.png]

7、你喜欢哪一类型的DIY手工艺制品?=(2aa4e3cfb024c7ff30a8846913966dfb1.png1)(68bbea9f5ae9ea8fccb12e5ca2209dcf.png

“漂亮女生”号称全国连锁店,相信他们有统一的进货渠道。店内到处贴着“10元以下任选”,价格便宜到令人心动。但是转念一想,发夹2.8元,发圈4.8元,皮夹子9.8元,好像和平日讨价还价杀来的心理价位也差不多,只不过把一只20元的发夹还到5元实在辛苦,现在明码标价倒也省心省力。92aa4e3cfb024c7ff30a8846913966dfb1.png1).

(二)对“碧芝”自制饰品店的分析【点评】本题第二个括号内有99个不同分母,不可能通分.这里采用的是先分母有理化,将分母化为整数,从而使每一项转化成两数之差,然后逐项相消.这种方法也叫做裂项相消法.

1)位置的优越性30.若xy为实数,且y9ca2cc686d258751a817a32fc8c97eef.png969479efaed492932427905e33259d70.png93b05c90d14a117ba52da1d743a43ab1.png.求d18e20bdd0c60849ed8f5b57f5f5de88.png855733fa578d4f272ac4f62df2622bca.png的值.

(四)大学生对手工艺制品消费的要求【提示】要使y有意义,必须满足什么条件?a2385ec1895b8d4e2475dd573c819dbd.png你能求出xy的值吗?98172fa5cd64b59e964682dac1504470.png

据介绍,经常光顾“碧芝”的都是些希望得到世界上“独一无二”饰品的年轻人,他们在琳琅满目的货架上挑选,然后亲手串连,他们就是偏爱这种DIY的方式,完全自助。【解】要使y有意义,必须9839c9c18932d1516788b480e19116f7.png,即d73061e115aac19e3bdca4bd064f9c39.png∴ xeca3bf81573307ec3002cf846390d363.png.当xeca3bf81573307ec3002cf846390d363.png时,y93b05c90d14a117ba52da1d743a43ab1.png

人民广场地铁站有一家名为“漂亮女生”的饰品店,小店新开,10平方米不到的店堂里挤满了穿着时尚的女孩子。不几日,在北京东路、淮海东路也发现了“漂亮女生”的踪影,生意也十分火爆。现在上海卖饰品的小店不计其数,大家都在叫生意难做,而“漂亮女生”却用自己独特的经营方式和魅力吸引了大批的女生。又∵ d18e20bdd0c60849ed8f5b57f5f5de88.png855733fa578d4f272ac4f62df2622bca.png317daca864fb3be6678e4d6798a18fd8.pnged8484446559e7e362220caa2aae6d23.png

1-5 购物是对消费环境的要求分布|7ea333e2ae014e95690364a9cde69077.png||6fb79d538b51cdd9621c5293011334e6.png|∵ xeca3bf81573307ec3002cf846390d363.pngy93b05c90d14a117ba52da1d743a43ab1.png,∴ a4df4c5aaf8f092e03a00f7109e5de74.pngaae37404f1219a49dafcaf015d5965f3.png

附件(二):∴ 原式=7ea333e2ae014e95690364a9cde69077.png15522dd73e3eb0bd7cc035fb02cc8a0f.png21a4ef11368714595850b3920a74f0509.pngxeca3bf81573307ec3002cf846390d363.pngy93b05c90d14a117ba52da1d743a43ab1.png时,

原式=2904dfcf20117290d58a9593c15037aad.pngd21848cdd835abcb491be1f151e9b6c6.png.【点评】解本题的关键是利用二次根式的意义求出x的值,进而求出y的值.

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