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Question

Giải các phương trình saua) x(x-1)+2x2-2=0b) 9x2-1=(3x+1)(2x-3)

Nguyễn Lê Phước Thịnh · Accepted Answer

a: $x\left(x-1\right)+2x^2-2=0$=>$x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)=0$=>$\left(x-1\right)\left(x+2x+2\right)=0$=>(x-1)(3x+2)=0=>$\left[{}\begin{matrix}x=1\x=-\dfrac{2}{3}\end{matrix}\right.$b: $9x^2-1=\left(3x+1\right)\left(2x-3\right)$=>$\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0$=>$\left(3x+1\right)\left(3x-1-2x+3\right)=0$=>(3x+1)(x+2)=0=>$\left[{}\begin{matrix}x=-\dfrac{1}{3}\x=-2\end{matrix}\right.$Câu trả lời: a: $x\left(x-1\right)+2x^2-2=0$=>$x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)=0$=>$\left(x-1\right)\left(x+2x+2\right)=0$=>(x-1)(3x+2)=0=>$\left[{}\begin{matrix}x=1\x=-\dfrac{2}{3}\end{matrix}\right.$b: $9x^2-1=\left(3x+1\right)\left(2x-3\right)$=>$\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0$=>$\left(3x+1\right)\left(3x-1-2x+3\right)=0$=>(3x+1)(x+2)=0=>$\left[{}\begin{matrix}x=-\dfrac{1}{3}\x=-2\end{matrix}\right.$

vovanquy · Answer

a: x(x−1)+2x2−2=0=>x(x−1)+2(x−1)(x+1)=0=>(x−1)(x+2x+2)=0=>(x-1)(3x+2)=0=>⎡⎣x=1x=−23b: 9x2−1=(3x+1)(2x−3)9=>(3x+1)(3x−1)−(3x+1)(2x−3)=0=>(3x+1)(3x−1−2x+3)=0=>(3x+1)(x+2)=0=>⎡⎣x=−13x=−2Câu trả lời: a: x(x−1)+2x2−2=0=>x(x−1)+2(x−1)(x+1)=0=>(x−1)(x+2x+2)=0=>(x-1)(3x+2)=0=>⎡⎣x=1x=−23b: 9x2−1=(3x+1)(2x−3)9=>(3x+1)(3x−1)−(3x+1)(2x−3)=0=>(3x+1)(3x−1−2x+3)=0=>(3x+1)(x+2)=0=>⎡⎣x=−13x=−2

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