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\(\left(3-2x\right)^2=\left(x-2\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-2\right)^2-\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)
\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)
\(\Leftrightarrow7x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)
`(3-2x)^2=(x-2)(2x-3)`
`<=>(2x-3)^2 -(x-2)(2x-3)=0`
`<=> (2x-3)(2x-3-x+2)=0`
`<=> (2x-3)(x-1)=0`
\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x+1)(x+3)=2x^2-2`
`<=>x^2+x+3x+3=2x^2-2`
`<=>x^2-4x-5=0`
`<=>x^2-5x+x-5=0`
`<=>x(x-5)+(x-5)=0`
`<=>(x-5)(x+1)=0`
`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$
Vậy `S={5,-1}`
Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2x^2+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3-2x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: S={-3;5}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow-x^2+6x-3+x^2-3x-1=0\)
\(\Leftrightarrow3x-4=0\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy: \(S=\left\{\frac{4}{3}\right\}\)
b) Ta có: \(3x\left(x-2\right)-x\left(1+3x\right)=14\)
\(\Leftrightarrow3x^2-6x-x-3x^2-14=0\)
\(\Leftrightarrow-7x-14=0\)
\(\Leftrightarrow-7x=14\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
hay x=-2
Vậy: S={-2}
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c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P=\(x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x=0\)
Vậy giá trị của P không phụ thuộc vào biến
Q=\(x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6=-8\)
Vậy giá trị của Q không phụ thuộc vào biến
Tick nha bạn 😘
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{x^2\left(x-2\right)+5\left(x-2\right)}{x-2}=x^2+5\)
(2x-1)(3-x)+(x-2)(x+3)=(1-x)(x-2)
<=> 6x-3-2x^2+x+x^2-2x+3x-6=x-x^2-2+2x
<=>-x^2+8x-9=-x^2+3x-2
<=>x^2-x^2+8x-3x=9-2
<=>5x=7
<=>x=7/5