Câu hỏi của Phạm Yến

Question

Tìm x

$\dfrac{x-3}{x+2}$=$\dfrac{x-5}{x+1}$

Đức Hiếu · Accepted Answer

$\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}$

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

$\Rightarrow x^2+x-3x-3=x^2-5x+2x-10$

$\Rightarrow x^2-2x-x^2+3x=-10+3$

$\Rightarrow x=-7$

Vậy...........

Chúc bạn học tốt!!!Câu trả lời: $\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}$

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

$\Rightarrow x^2+x-3x-3=x^2-5x+2x-10$

$\Rightarrow x^2-2x-x^2+3x=-10+3$

$\Rightarrow x=-7$

Vậy...........

Chúc bạn học tốt!!!

Mashiro Shiina · Answer

$\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}$

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

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

$\Rightarrow x^2+x-3x-3=x^2+2x-5x-10$

$\Rightarrow x^2-2x-3=x^2-3x-10$

$\Rightarrow x^2-2x+3x=x^2-10+3$

$\Rightarrow x^2+x=x^2-7$

$\Rightarrow x=-7$Câu trả lời: $\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}$

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

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

$\Rightarrow x^2+x-3x-3=x^2+2x-5x-10$

$\Rightarrow x^2-2x-3=x^2-3x-10$

$\Rightarrow x^2-2x+3x=x^2-10+3$

$\Rightarrow x^2+x=x^2-7$

$\Rightarrow x=-7$

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