Câu hỏi của Bảo Ngọc cute

Question

Tìm x

$\left(x-5\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2$

Linh Nguyễn · Accepted Answer

$\left(x-5\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2$

$\Rightarrow\left[x\left(x+4\right)-5\left(x+4\right)\right]-\left[x\left(x+3\right)-1\left(x+3\right)\right]=-2x^2$

$\Rightarrow\left(x^2+4x-5x-20\right)-\left(x^2+3x-x-3\right)=-2x^2$

$\Rightarrow\left(x^2-x-20\right)-\left(x^2+2x-3\right)=-2x^2$

$\Rightarrow x^2-x-20-x^2-2x+3=-2x^2$

$\Rightarrow-x-17=-2x^2$

Cạn....Câu trả lời: $\left(x-5\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2$

$\Rightarrow\left[x\left(x+4\right)-5\left(x+4\right)\right]-\left[x\left(x+3\right)-1\left(x+3\right)\right]=-2x^2$

$\Rightarrow\left(x^2+4x-5x-20\right)-\left(x^2+3x-x-3\right)=-2x^2$

$\Rightarrow\left(x^2-x-20\right)-\left(x^2+2x-3\right)=-2x^2$

$\Rightarrow x^2-x-20-x^2-2x+3=-2x^2$

$\Rightarrow-x-17=-2x^2$

Cạn....

Violympic toán 8