Câu hỏi của Minh Thảo

Question

tìm x:

(x2-3)2 - (8x-6)(x2-3)+(3-4x)2 =0

Phong Thần · Accepted Answer

$\left(x^2-3\right)^2-\left(8x-6\right)\left(x^2-3\right)+\left(3-4x\right)^2=0$

$\Rightarrow\left(x^2-3\right)^2-2\left(4x-3\right)\left(x^2-3\right)+\left(3-4x\right)^2=0$

$\Rightarrow\left(x^2-3\right)^2+2\left(3-4x\right)\left(x^2-3\right)+\left(3-4x\right)^2=0$

$\Rightarrow\left(x^2-3+3-4x\right)^2=0$

$\Rightarrow\left(x^2-4x\right)^2=0$

$\Rightarrow x^2-4x=0$

$\Rightarrow x\left(x-4\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=0\x-4=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=0\x=4\end{matrix}\right.$Câu trả lời: $\left(x^2-3\right)^2-\left(8x-6\right)\left(x^2-3\right)+\left(3-4x\right)^2=0$