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a) \(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\cdot5}{5}=2\)
b)\(\frac{3}{8}=\frac{6}{x}\Rightarrow x=\frac{6\cdot8}{3}=16\)
c)\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{27}{9}=3\)
d) \(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4\cdot6}{8}=3\)
e)
\(\frac{3}{x-5}=-\frac{4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow3x+6+4x-20=0\\ \Leftrightarrow7x-14=0\\ \Leftrightarrow x=2\)
\(g,\frac{x}{-2}=-\frac{8}{x}\Rightarrow x^2=16\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy có 2 giá trị của x là 4 ; -4
a) Ta có: \(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{6\cdot8}{3}=\frac{48}{3}=16\)
Vậy: x=16
c) Ta có: \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\cdot27}{9}=\frac{27}{9}=3\)
Vậy: x=3
d) Ta có: \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\cdot6}{8}=\frac{24}{8}=3\)
Vậy: x=3
e) Ta có: \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\cdot\left(x+2\right)=-4\cdot\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+6+4x-20=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\)
hay x=2
Vậy: x=2
g) Sửa đề: \(\frac{x}{-2}=\frac{-8}{x}\)
Ta có: \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=\left(-8\right)\cdot\left(-2\right)=16\)
hay x∈{4;-4}
Vậy: x∈{4;-4}
mk muốn xem bài của mk đúng hay sai thôi !
chứ làm thì mk làm xong rồi !
a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)
\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow x-2=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)
\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)
hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)
Vậy: \(x=\frac{13}{15}\)
c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)
\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)
\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)
\(\Leftrightarrow3x=-160\)
hay \(x=\frac{-160}{3}\)
Vậy: \(x=\frac{-160}{3}\)
d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)
a/ (x - 2)3 + \(\frac{8}{27}\) = 0
=> (x - 2)3 = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)
=> x - 2 = \(-\frac{2}{3}\)
=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)
b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)
=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)
=> \(x=\frac{13}{60}.4=\frac{13}{15}\)
c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)
=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)
=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)
=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)
a) \(\frac{1}{9}=\frac{x}{27}\)
\(\Rightarrow x=\frac{1}{9}\cdot27\)
\(\Rightarrow x=3\)
b) \(\frac{4}{x}=\frac{8}{6}\)
\(\Rightarrow x=4:\frac{8}{6}\)
\(\Rightarrow x=3\)
c) \(\frac{x}{3}-\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\Rightarrow x=\frac{7}{10}\cdot3\)
\(\Rightarrow x=\frac{21}{10}=2,1\)
\(a,\frac{1}{9}\)=\(\frac{3}{27}\)
\(b,\frac{4}{3}\)=\(\frac{8}{6}\)
\(c,\frac{x}{3}\)-\(\frac{1}{2}=\frac{1}{5}\)
\(\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(\)