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V
8 tháng 10 2019
ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)
5 tháng 8 2018
a) \(3x\left(x-2\right)-4x+8=0\)
\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)
b) \(3\left(2x-1\right)^2+2-4x=0\)
\(\Leftrightarrow3\left(2x-1\right)^2-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(6x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\6x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{6}\end{cases}}\)
14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
QN
4
1 tháng 9 2018
a) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=0\)
\(\Leftrightarrow5x+8=8\)
\(\Leftrightarrow5x=8-8\)
\(\Leftrightarrow x=5.0\)
\(\Leftrightarrow x=0\)
b)
11 tháng 10 2020
a) \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;3\right\}\)
b) \(\left(x+8\right)^2=121\)
\(\Leftrightarrow\left(x+8\right)^2-121=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+19\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{-19;3\right\}\)
c) \(x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{0;2\right\}\)
d) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
13 tháng 10 2017
a. \(x^3+4x^2-31x-70=x^3+2x^2+2x^2+4x-35x-70=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-\left(35x+70\right)=x^2\left(x+2\right)+2x\left(x+2\right)-35\left(x+2\right)=\left(x+2\right)\left(x^2+2x-35\right)=\left(x+2\right)\left(x^2+7x-5x-35\right)=\left(x+2\right)[\left(x^2+7x\right)-\left(5x+35\right)]=\left(x+2\right)\left[x\left(x+7\right)-5\left(x+7\right)\right]=\left(x+2\right)\left(x+7\right)\left(x-5\right)\)
b. \(y^2-y-12=y^2-4y+3y-12=\left(y^2+3y\right)-\left(4y+12\right)=y\left(y+3\right)-4\left(x+3\right)=\left(x+3\right)\left(y-4\right)\)
c. \(x^2-3x^2+4x-2=-2x^2+4x-2=-2\left(x^2-2x+1\right)=-2\left(x-1\right)^2\)