Question

Tinh gia tri cua bieu thuc A = x⁶-2*x⁴+x³+x²-x biet x³-x=8?

Answer 1

Ta có: \(A=x^6-2x^4+x^3+x^2-x\)

\(\Rightarrow A=\left(x^6-2x^4+x^2\right)+\left(x^3-x\right)\)

\(\Rightarrow A=\left[\left(x^3\right)^2-2x^3x+x^2\right]+\left(x^3-x\right)\)

\(\Rightarrow A=\left(x^3-x\right)^2+\left(x^3-x\right)\)\(\left(1\right)\)

Thay \(x^3-x=8\)vào \(\left(1\right)\)ta có:

\(\Rightarrow A=8^2+8=72\)

Vậy \(A=72\)

Answer 2

A=x^6-2x^4+x^2+(x^3-x)

=x^6-x^4-x^4+x^2+(x^3-x)

=x^3(x^3-x)-x(x^3-x)+(x^3-x)

=(x^3-x)(x^3-x)+(x^3-x)=8.8+8=8*9=72