1. biết x2-2y2=xy,y\(\ne\)0,x+y\(\ne\)0. thì gia tri cua bieu thuc Q=\(\frac{x+y}{x-y}\)=2.cho x\(\ne\)0,y\(\ne\)0 thoa man x+y=4 ;xy=2 .gia tri cua bieu thuc A=\(\frac{1}{x^3}+\frac{1}{y^3}\)la3.gia tri cua bieu thuc...
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1. biết x2-2y2=xy,y\(\ne\)0,x+y\(\ne\)0. thì gia tri cua bieu thuc Q=\(\frac{x+y}{x-y}\)=
2.cho x\(\ne\)0,y\(\ne\)0 thoa man x+y=4 ;xy=2 .gia tri cua bieu thuc A=\(\frac{1}{x^3}+\frac{1}{y^3}\)la
3.gia tri cua bieu thuc A=\(\frac{81^8-1}{\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}\)la
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Bài 3:
Ta có:
\(81^8-1=\left(9^2\right)^8-1=\left[\left(3^2\right)^2\right]^8-1=3^{32}-1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Do đó:
\(A=3^4-1=80\)