题目内容
①若
=
,则
= .
②若
=
=
,则
= .
③已知
+
=4,则
= .
④若m+n=5,mn=3,则
+
= .
| m |
| n |
| 5 |
| 3 |
| m2+2mn+n2 |
| m2-mn-2n2 |
②若
| x |
| 3 |
| y |
| 4 |
| z |
| 5 |
| x2-3xy+2z2 |
| 3x2+2xy-z2 |
③已知
| 1 |
| a |
| 1 |
| b |
| 4a+3ab+4b |
| -3a+2ab-3b |
④若m+n=5,mn=3,则
| n |
| m |
| m |
| n |
考点:分式的化简求值
专题:计算题
分析:①对所要求的式子进行变形,即分子和分母都除以式子n2,然后把条件代入即可求值;
②令
=
=
=k,则x=3k,y=4k,z=5k,然后代入即可求值;
③由条件可以得到a+b=4ab,然后代入进行求值即可;
④把要求的式子进行变形为
,然后把条件代入即可求值.
②令
| x |
| 3 |
| y |
| 4 |
| z |
| 5 |
③由条件可以得到a+b=4ab,然后代入进行求值即可;
④把要求的式子进行变形为
| (m+n)2-2mn |
| mn |
解答:解:①
=
=
=-8;
②令
=
=
=k,则x=3k,y=4k,z=5k,
所以
=
=
=-
;
③由
+
=4得a+b=4ab,
所以
=
=
=
=-
;
④
+
=
=
=
=
.
| m2+2mn+n2 |
| m2-mn-2n2 |
(
| ||||
(
|
(
| ||||
(
|
②令
| x |
| 3 |
| y |
| 4 |
| z |
| 5 |
所以
| x2-3xy+2z2 |
| 3x2+2xy-z2 |
| (3k)2-3×3k×4k+2×(5k)2 |
| 3×(3k)2+2×3x×4k-(5k)2 |
| -5k2 |
| 26k2 |
| 5 |
| 26 |
③由
| 1 |
| a |
| 1 |
| b |
所以
| 4ab+3ab+4b |
| -3a+2ab-3b |
| 4(a+b)+3ab |
| -3(a+b)+2ab |
| 4×4ab+3ab |
| -3×4ab+2ab |
| 19ab |
| -10ab |
| 19 |
| 10 |
④
| n |
| m |
| m |
| n |
| n2+m2 |
| mn |
| (m+n)2-2mn |
| mn |
| 52-2×3 |
| 3 |
| 19 |
| 3 |
点评:本题考查的是分式的化简求值,熟知分式混合运算的法则是解答此题的关键.
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