Câu hỏi của Trần Đức Gia Khánh

Question

1) 2x + x (x - 5) = 3x^2 -x

2) 15 - 5 ( 1 - 2x) = 12 - x

3) 2/3 - 1/3 ( x-3/2 ) - 1/2 ( 2x +1) =5

4) | x - 4/5 | = 3/5

Nguyễn Lê Phước Thịnh · Accepted Answer

1: Ta có: $2x+x\left(x-5\right)=3x^2-x$

$\Leftrightarrow2x+x^2-5x-3x^2+x=0$

$\Leftrightarrow-2x^2-2x=0$

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

Vì -2≠0

nên $\left[{}\begin{matrix}x=0\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

Vậy: x∈{0;-1}

2) Ta có: $15-5\left(1-2x\right)=12-x$
$\Leftrightarrow15-5+10x-12+x=0$

$\Leftrightarrow11x-2=0$

$\Leftrightarrow11x=2$

hay $x=\frac{2}{11}$

Vậy: $x=\frac{2}{11}$

3) Ta có: $\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5$

$\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0$

$\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0$

$\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}$

hay $x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}$

Vậy: $x=\frac{-13}{4}$

4) Ta có: $\left|x-\frac{4}{5}\right|=\frac{3}{5}$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\x=\frac{1}{5}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{1}{5};\frac{7}{5}\right\}$Câu trả lời: 1: Ta có: $2x+x\left(x-5\right)=3x^2-x$

$\Leftrightarrow2x+x^2-5x-3x^2+x=0$

$\Leftrightarrow-2x^2-2x=0$

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

Vì -2≠0

nên $\left[{}\begin{matrix}x=0\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

Vậy: x∈{0;-1}

2) Ta có: $15-5\left(1-2x\right)=12-x$
$\Leftrightarrow15-5+10x-12+x=0$

$\Leftrightarrow11x-2=0$

$\Leftrightarrow11x=2$

hay $x=\frac{2}{11}$

Vậy: $x=\frac{2}{11}$

3) Ta có: $\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5$

$\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0$

$\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0$

$\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}$

hay $x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}$

Vậy: $x=\frac{-13}{4}$

4) Ta có: $\left|x-\frac{4}{5}\right|=\frac{3}{5}$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\x=\frac{1}{5}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{1}{5};\frac{7}{5}\right\}$

Bùi Lan Anh · Answer

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

$\Leftrightarrow2x+x^2-5x=3x^2-x$

$\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0$

$\Leftrightarrow-2x-2x^2=0$

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

$\Leftrightarrow\left[{}\begin{matrix}-2x=0\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

2. $15-5\left(1-2x\right)=12-x$

$\Leftrightarrow15-5+10x=12-x$

$\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0$

$\Leftrightarrow-2+11x=0$

$\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}$

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

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

$\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0$

$\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0$

$\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}$

4. $\left|x-\frac{4}{5}\right|=\frac{3}{5}$

$\Rightarrow x-\frac{4}{5}=-\frac{3}{5}$ hoặc $x-\frac{4}{5}=\frac{3}{5}$

$TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}$

$TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}$Câu trả lời: 1. $2x+x\left(x-5\right)=3x^2-x$