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\(x^2=\frac{144}{25}=\left(\frac{12}{5}\right)^2\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{5}\\x=-\frac{12}{5}\end{cases}}\)
\(\frac{x}{3,15}=\frac{0,15}{7,2}\)
\(\Rightarrow x=\frac{0,15.3,15}{7,2}\)
\(\Rightarrow x=0,065625\)
b) \(\frac{-2,6}{x}=\frac{-12}{42}\)
\(\Rightarrow x=\frac{-2,6.42}{-12}\)
\(\Rightarrow x=9,1\)
Ta có:
\(\frac{x^2}{6}=\frac{24}{25}\Leftrightarrow x^2.25=24.6\)
\(\Leftrightarrow x^2=\frac{24.6}{25}=\frac{144}{25}=5,76\)
\(\Leftrightarrow x=\sqrt{5,76}=2,4;x=-\sqrt{5,76}=-2,4\)
Vậy \(x=2,4\) hoặc \(x=-2,4\)
a)x-2/5=3/8
(x-2).8=5.3
x-2=5.3:8
x-2=1
=>x=3
b,c )tương tự
a) ĐKXĐ: \(x\ne-5\)
\(\Leftrightarrow7x-7=6x+30\\ \Leftrightarrow x=37\)
b) \(\Leftrightarrow25x^2=144\\ \Leftrightarrow x^2=\dfrac{144}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
a)
b) \(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Leftrightarrow\left(5x\right)^2=144\)
\(\Leftrightarrow\left(5x\right)^2=12^2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2\cdot25=6\cdot24\)
\(x^2\cdot25=144\)
\(x^2=144:25\)
\(x=\pm\sqrt{\frac{12}{5}}\)
\(\frac{x.x}{6}=\frac{24}{25}\Rightarrow25.x^2=24.6=144\Rightarrow x^2=\frac{144}{25}=\left(\frac{12}{5}\right)^2=\left(-\frac{12}{5}\right)^2\)
=> x =12/5 hoặc x = -12/5
\(x\times\frac{x}{6}=\frac{24}{25}\)
\(x=\frac{24}{25}:\frac{x}{6}\)
\(x=\frac{84}{25x}\)
\(25x^2=84\)
\(x^2=3,36\)
\(x=\frac{2\sqrt{21}}{5}\)