TÌM X BIẾT :|3x-5| - |x+2| = 0
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a,\(\left(x-1\right)^2-\left(2x\right)^2=0< =>\left(x-1-2x\right)\left(x-1+2x\right)=0\)
\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)
b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)
\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)
a, \(\left(x-1\right)^2-\left(2x\right)^2=0\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)
b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x-5-x\right)=0\Leftrightarrow\left(3x-5\right)\left(2x-5\right)=0\Leftrightarrow x=\frac{5}{3};x=\frac{5}{2}\)
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a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
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x2( x + 1 ) + 2x( x + 1 ) = 0 <=> x( x + 1 )( x + 2 ) = 0 <=> x = 0 hoặc x = -1 hoặc x = -2
x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5
Trả lời:
1, \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=0;x=-1;x=-2\)
Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.
2, \(x\left(3x-1\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5\end{cases}}}\)
Vậy x = 1/3; x = - 5 là nghiệm của pt.
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\(\Leftrightarrow\left(x+1\right)\left(3x-5\right)=0\)
hay \(x\in\left\{-1;\dfrac{5}{3}\right\}\)
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a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
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a) 5x^2 + 15x = 0
=> 5x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3
@Hoàng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x-\left(\dfrac{2}{5}+x\right)=0,4\)
\(3x-\dfrac{2}{5}-x=0,4\)
\(2x-\dfrac{2}{5}=\dfrac{2}{5}\)
\(2x=\dfrac{2}{5}+\dfrac{2}{5}\)
\(2x=\dfrac{4}{5}\)
\(x=\dfrac{2}{5}\)
Ta có: \(3x-\left(x+\dfrac{2}{5}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow2x-\dfrac{2}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow2x=\dfrac{4}{5}\)
hay \(x=\dfrac{2}{5}\)