(2x-3)^2=4/25
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tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
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g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)
\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)
a: \(=\dfrac{2x^3-3x^2+4x^2-6x-2x+3}{2x-3}=x^2+2x-1\)
b: \(=\dfrac{x-5}{\left(x-2\right)^2}\cdot\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{2}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\2x-3=-\dfrac{2}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\2x=\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Leftrightarrow\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\-2x+3=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\-2x=-\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
Vậy...