Tìm x biết : (2x - 4)^2 : 3/4 - 1/3 = 1
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a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
\(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(3x=\frac{3}{5}-\frac{3}{4}=-\frac{3}{20}\)
\(x=-\frac{1}{20}\)
b) \(\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
=> \(\orbr{\begin{cases}\frac{x}{2}-\frac{1}{3}=\frac{7}{4}\\\frac{x}{2}-\frac{1}{3}=-\frac{7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}}\)
a, \(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(-2x-\frac{3}{4}-x+\frac{3}{5}=0\)
\(-3x-\frac{3}{20}=0\)
\(\frac{3}{20}=-3x\Leftrightarrow x=\frac{1}{20}\)
b: =>2x=16/2=8
=>x=4
a: Sửa đề: (3/2)^2x-1=(3/2)^5x-4
=>2x-1=5x-4
=>-3x=-3
=>x=1
Trả lời:
\(\left(2x-4\right)^2:\frac{3}{4}-\frac{1}{3}=1\)
\(\Leftrightarrow\left(2x-4\right)^2:\frac{3}{4}=\frac{4}{3}\)
\(\Leftrightarrow\left(2x-4\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4=1\\2x-4=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}}\)
Vậy x = 5/2; x = 3/2
\(\left(2x-4\right)^2\div\frac{3}{4}=\frac{4}{3}\)
\(\left(2x-4\right)^2=1\)
\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=-1\\2x-4=1\end{cases}\Rightarrow\orbr{\begin{cases}2x=3\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{5}{2}\end{cases}}}\)