a) (3x-2):\(1\frac{2}{5}\)=\(2\frac{3}{7}\)
(3x-2):\(\frac{7}{5}\)=\(\frac{17}{7}\)
3x-2 =\(\frac{17}{7}.\frac{7}{5}\)
3x-2 =\(\frac{17}{5}\)
3x =\(\frac{17}{5}+2\)
3x =\(\frac{27}{5}\)
x =\(\frac{27}{5}:3\)
x =\(\frac{27}{5}.\frac{1}{3}\)
x =\(\frac{9}{5}\)
b) x:0,16=9:x
x.x =9.0,16
x2 =\(\frac{36}{25}\)
x2 =\(\left(\frac{6}{5}\right)^2\)
⇒x =\(\frac{6}{5}\)
a) Ta có: \(\left(3x-2\right):1\frac{2}{5}=2\frac{3}{7}:2\frac{3}{5}\)
\(\Leftrightarrow\left(3x-2\right):\frac{7}{5}=\frac{17}{7}:\frac{13}{5}\)
\(\Leftrightarrow\left(3x-2\right)\cdot\frac{5}{7}=\frac{17}{7}\cdot\frac{5}{13}=\frac{85}{91}\)
\(\Leftrightarrow3x-2=\frac{85}{91}:\frac{5}{7}=\frac{85}{91}\cdot\frac{7}{5}=\frac{17}{13}\)
\(\Leftrightarrow3x=\frac{17}{13}+2=\frac{43}{13}\)
\(\Leftrightarrow x=\frac{43}{13}:3=\frac{43}{13}\cdot\frac{1}{3}\)
hay \(x=\frac{43}{39}\)
Vậy: \(x=\frac{43}{39}\)
b) ĐKXĐ: x≠0
Ta có: \(\frac{x}{0.16}=\frac{9}{x}\)
\(\Leftrightarrow x\cdot x=9\cdot0.16\)
\(\Leftrightarrow x^2=1.44\)
hay \(x\in\left\{1.2;-1.2\right\}\)(nhận)
Vậy: \(x\in\left\{1.2;-1.2\right\}\)
