\(\frac{x}{-2}=\frac{-8}{x}tìmx\)
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ta có \(x+1.x+1=8.2=16\)
\(x+1.x+1=4^2\)
\(x+1=4\)
\(x=3\)
Theo đầu bài ta có:
\(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=2\cdot8\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow x+1=\sqrt{16}\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{7}{8}.\frac{64}{49}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{8}{7}\)
\(\frac{3}{2}x=\frac{8}{7}+\frac{11}{5}=\frac{117}{35}\)
\(x=\frac{117}{35}:\frac{3}{2}=\frac{78}{35}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{7}{8}.\frac{64}{49}\)
\(\frac{3}{2}x-\frac{11}{5}=\frac{8}{7}\)
\(\frac{3}{2}x=\frac{8}{7}+\frac{11}{5}\)
\(\frac{3}{2}x=\frac{117}{35}\)
\(x=\frac{117}{35}:\frac{3}{2}\)
\(x=\frac{78}{35}\)
\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
\(\frac{-2}{3}.x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{-2}{3}=-\frac{3}{20}\)
\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(x+3\frac{1}{7}\right)+7\frac{1}{2}=1\frac{69}{86}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\left(\frac{14}{6}+\frac{21}{6}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{15}{2}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{645}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{-245}{43}\)
\(x+\frac{22}{7}=\frac{35}{6}:\frac{-245}{43}=\frac{35}{6}\cdot\frac{-43}{245}\)
\(x+\frac{22}{7}=\frac{-43}{42}\)
\(x=\frac{-43}{42}-\frac{22}{7}=\frac{-43}{42}-\frac{132}{42}\)
\(x=\frac{-25}{6}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)
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:
\(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x.x=\left(-2\right).\left(-8\right)\Rightarrow x^2=16=4^4\Rightarrow x=4\)