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\(\frac{3}{2}x-\frac{11}{5}=\frac{7}{8}.\frac{64}{49}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{8}{7}\)
\(\frac{3}{2}x=\frac{8}{7}+\frac{11}{5}=\frac{117}{35}\)
\(x=\frac{117}{35}:\frac{3}{2}=\frac{78}{35}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{7}{8}.\frac{64}{49}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{8}{7}\)
\(\frac{3}{2}x=\frac{8}{7}+\frac{11}{5}\)
\(\frac{3}{2}x=\frac{117}{35}\)
\(x=\frac{117}{35}:\frac{3}{2}\)
\(x=\frac{78}{35}\)
\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
\(\frac{-2}{3}.x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{-2}{3}=-\frac{3}{20}\)
\(\frac{5}{7}:x+\frac{4}{5}=\frac{1}{6}\)
\(\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}=-\frac{19}{30}\)
\(x=\frac{5}{7}:\frac{-19}{30}=-\frac{150}{133}\)
\(\frac{5}{7}:x+\frac{4}{5}=\frac{1}{6}\Rightarrow\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}=-\frac{19}{30}\Rightarrow x=\frac{5}{7}:-\frac{19}{30}=\frac{5}{7}.-\frac{30}{19}=-\frac{150}{133}\)
a/ ĐKXĐ: \(x\ne-1\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x=x+1\)
\(\Rightarrow29x=11\Rightarrow x=\frac{11}{29}\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow1-\left(\sqrt{x}-2\right)=3-\sqrt{x}\)
\(\Leftrightarrow3=3\) (luôn đúng)
Vậy nghiệm của pt là \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne7\)
\(\Leftrightarrow8-x-8\left(x-7\right)=1\)
\(\Leftrightarrow8-x-8x+56=1\)
\(\Leftrightarrow-9x=-63\Rightarrow x=7\left(ktm\right)\)
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ne4\)
\(\Leftrightarrow\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}=\frac{-9}{6\left(x-4\right)}-\frac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
\(\Rightarrow x=5\)
e/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)
Áp dụng kiến thức trên ta có:
\(\frac{x}{-2}=-\frac{8}{x}\Leftrightarrow x.x=\left(-8\right).\left(-2\right)=16\)
\(\Leftrightarrow x^2=16\rightarrow x\in\left\{-4;4\right\}\)
Ta có:
\(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x.x=\left(-2\right).\left(-8\right)\Rightarrow x^2=16=4^4\Rightarrow x=4\)