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13 tháng 9 2021
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
11 tháng 3 2023
`a,4x-10=0 `
`<=> 4x=10`
`<=>x=10/4`
`<=>x=5/2`
`b, 7-3x=9-x `
`<=>-3x+x=9-7`
`<=>-2x=2`
`<=>x=-1`
`c, 2x-(3-5x) = 4(x+3)`
`<=>2x-3+5x=4x+12`
`<=>2x+5x-4x=12+3`
`<=>3x=15`
`<=>x=5`
`d, 5-(6-x)=4(3-2x) `
`<=>5-6+x=12-8x`
`<=>x+8x=12-5+6`
`<=>9x=13`
`<=>x=13/9`
`e, 4(x+3)=-7x+17 `
`<=>4x+12=-7x+17`
`<=>4x+7x=17-12`
`<=>11x=5`
`<=>x=5/11`
`f, 5(x-3) - 4=2(x-1)+7`
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`g, 5(x-3)-4=2(x-1)+7 `
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`h,4(3x-2)-3(x-4)=7x+20`
`<=>12x-8-3x+12=7x+20`
`<=>12x-3x-7x=20+8+12`
`<=>2x=40`
`<=>x=20`
4 tháng 6 2022
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
28 tháng 7 2021
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
28 tháng 7 2021
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
\(\dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)-\dfrac{17}{3}=2\dfrac{5}{3}?\)
\(\dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)-\dfrac{17}{3}=\dfrac{11}{3}\\ \dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)=\dfrac{11}{3}+\dfrac{17}{3}=\dfrac{28}{3}\\ x-\dfrac{5}{3}=\dfrac{3}{2}:\dfrac{28}{3}=\dfrac{9}{56}\\ x=\dfrac{9}{56}+\dfrac{5}{3}\\ x=\dfrac{307}{168}?\)
307/168