① 4(m+2n)²-9(2n-m)² 因式分解②已知(a²+b²)(a²+b²-8)+16=0,求a²+b²的 值
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① 4(m+2n)²-9(2n-m)² 因式分解②已知(a²+b²)(a²+b²-8)+16=0,求a²+b²的 值
① 4(m+2n)²-9(2n-m)² 因式分解
②已知(a²+b²)(a²+b²-8)+16=0,求a²+b²的 值
① 4(m+2n)²-9(2n-m)² 因式分解②已知(a²+b²)(a²+b²-8)+16=0,求a²+b²的 值
4(m+2n)²-9(2n-m)²
=(2m+4n)^2-(6n-3m)^2
=(2m+4n+6n-3m)(2m+4n-6n+3m)
=(10n-m)(5m-2n)
(a²+b²)(a²+b²-8)+16=0
设a^2+b^2=t,则t>=0
t(t-8)+16=0
t^2-8t+16=0
(t-4)^2=0
t=4
即a^2+b^2=4
① 4(m+2n)²-9(2n-m)²
=[2(m+2n)]²-[3(2n-m)]²
=(2m+4n+6n-3m)(2m+4n-6n+3m)
=(10n-m)(5m-2n)
(a²+b²)(a²+b²-8)+16=0
(a²+b²)²-8(a...
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① 4(m+2n)²-9(2n-m)²
=[2(m+2n)]²-[3(2n-m)]²
=(2m+4n+6n-3m)(2m+4n-6n+3m)
=(10n-m)(5m-2n)
(a²+b²)(a²+b²-8)+16=0
(a²+b²)²-8(a²+b²)+16=0
(a²+b²-4)²=0
所以a²+b²-4=0
a²+b²=4
收起
将4(m+2n)²看成【2(m+2n)】² 9(2n-m)²看成【3(2n-m)】² 然后用平方差
得(10n-m)(5m-2n)
将a²+b²看成整体 (x) 化为x(x-8)+16=0 打开得完全平方式(x-4)²=0 得a²+b²=x=4