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`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
\(\frac{8}{5}=\frac{-12}{x}\left(x\ne0\right)\)\(\Leftrightarrow8x=-60\)\(\Leftrightarrow x=\frac{-60}{8}=\frac{-15}{2}\)(tmđk)
\(\frac{x-1}{-4}=\frac{-4}{x-1}\left(x\ne1\right)\)\(\Leftrightarrow\left(x-1\right)^2=16\)\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)
\(\frac{8}{5}=\frac{-12}{x}\)
\(\Rightarrow8x=-60\)
\(x=-60:8\)
\(x=-7,5\)
Vậy x=-7,5
\(\frac{x-1}{-4}=\frac{-4}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\left(x-1\right)^2=4^2\)
\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4+1=5\\x=-4+1=-3\end{cases}}\)
vậy x=5 hoặc x=-3
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)
=>x=-3; y=28; z=-60
b: 5/12=x/-72
=>x=-72*5/12=-6*5=-30
c: =>x+3=-5
=>x=-8
\(a)\) \(-\left(x+84\right)+213=-16\)
\(\Leftrightarrow\)\(-x-84+213=-16\)
\(\Leftrightarrow\)\(x=213-84+16\)
\(\Leftrightarrow\)\(x=145\)
Vậy \(x=145\)
\(b)\) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=2\)
Chúc bạn học tốt ~
a) \(-\left(x+84\right)+213=-16\)
\(-\left(x+84\right)=-16-213\)
\(-\left(x+84\right)=-229\)
\(\Rightarrow x+84=229\)
\(\Rightarrow x=229-84=145\)
Vậy \(x=145\)
b) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|\frac{-1}{4}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|-1\right|\)
\(\left(x-1\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1+1=2\\x=-1+1=0\end{cases}}\)
Vậy \(x\in\left\{0;2\right\}\)
\(a)\frac{-84}{14}< 3x< \frac{108}{9}\)
\(\Rightarrow-6< 3x< 12\)
\(\Rightarrow-2< x< 4\)
\(\Rightarrow x=3\)
\(c)\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\cdot9\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=25\)
\(d)\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow(-x)\cdot x=(-9)\cdot4\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x=\pm6\)
\(e)\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x(x+1)=18\cdot4\)
\(\Rightarrow x(x+1)=72\)
\(\Rightarrow x(x+1)=8\cdot9=(-9)\cdot(-8)\)
Do đó : x = 8 hoặc x = -9
a) \(4^x+4^{x+1}=80\)
\(\Rightarrow4^x+4^x.4=80\)
\(\Rightarrow4^x.\left(1+4\right)=80\)
\(\Rightarrow4^x.5=80\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
Vậy x = 2