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Question

Tìm x biết

a) $(x-2)(x+1)+(x-2)(3-x)=0$

b) $\dfrac{x^2+1}{x^2+3}.\dfrac{(2x-5)(x^2+3)}{x^2+1}=5$

Sáng · Accepted Answer

a. $\left(x-2\right)\left(x+1\right)+\left(x-2\right)\left(3-x\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x+1+3-x\right)=0$

$\Leftrightarrow4\left(x-2\right)=0\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

b. $\dfrac{x^2+1}{x^3+3}.\dfrac{\left(2x-5\right)\left(x^2+3\right)}{x^2+1}=5$

$\Leftrightarrow\dfrac{\left(x^2+1\right)\left(2x-5\right)\left(x^2+3\right)}{\left(x^2+3\right)\left(x^2+1\right)}=5$

$\Leftrightarrow2x-5=5\Leftrightarrow x=5$Câu trả lời: a. $\left(x-2\right)\left(x+1\right)+\left(x-2\right)\left(3-x\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x+1+3-x\right)=0$

$\Leftrightarrow4\left(x-2\right)=0\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

b. $\dfrac{x^2+1}{x^3+3}.\dfrac{\left(2x-5\right)\left(x^2+3\right)}{x^2+1}=5$

$\Leftrightarrow\dfrac{\left(x^2+1\right)\left(2x-5\right)\left(x^2+3\right)}{\left(x^2+3\right)\left(x^2+1\right)}=5$

$\Leftrightarrow2x-5=5\Leftrightarrow x=5$

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