tìm x thuộcZ biết
x-2/4=9/x-2
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\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?
\(\dfrac{x^2}{20}=\dfrac{4}{5}\)
\(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)
\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)
\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)
\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)
\(\Leftrightarrow xy=63\)
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)
\(\Leftrightarrow\left(x+2\right)^3=343\)
=>x+2=7
hay x=5
\(\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=9.4\)
\(\left(x-2\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)
\(\frac{x-2}{4}=\frac{9}{x-2}\)
<=>(x-2)(x-2)=9.4
<=>(x-2)2=36
<=>(x-2)2=62
=>x-2=-6 hoặc x-2=6
x=-6+2 x=6+2
x=-4 x=8
Vậy x\(\in\){-4;8}