题目内容
(1)已知A=2x+y,B=2x-y,计算A2-B2;
(2)已知
=
=
,求
.
(2)已知
| x |
| 2 |
| y |
| 3 |
| z |
| 4 |
| xy+yz+zx |
| x2+y2+z2 |
分析:(1)直接把A=2x+y,B=2x-y代入所求代数式进行计算即可;
(2)令
=
=
=k(k≠0),则x=2k,y=3k,z=4k,再代入代数式尽心计算即可.
(2)令
| x |
| 2 |
| y |
| 3 |
| z |
| 4 |
解答:解:(1)∵A=2x+y,B=2x-y,
∴A2-B2=(2x+y)2-(2x-y)2
=(4x2+4xy+y2)-(4x2-4xy+y2)
=4x2+4xy+y2-4x2+4xy-y2
=8xy;
(2)令
=
=
=k(k≠0),则x=2k,y=3k,z=4k,
故原式=
=
=
=
.
∴A2-B2=(2x+y)2-(2x-y)2
=(4x2+4xy+y2)-(4x2-4xy+y2)
=4x2+4xy+y2-4x2+4xy-y2
=8xy;
(2)令
| x |
| 2 |
| y |
| 3 |
| z |
| 4 |
故原式=
| 2k•3k+3k•4k+4k•2k |
| (2k)2+(3k)2+(4k)2 |
=
| 6k2+12k2+8k2 |
| 4k2+9k2+16k2 |
=
| 26k2 |
| 29k2 |
=
| 26 |
| 29 |
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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