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a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
\(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Rightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Rightarrow x=-\frac{87}{140}\)
tíc mình nha
a) Ta có: \(\left(x-1\right)^2\ge\)0 \(\forall\)x
\(\left|y+2\right|\ge0\)\(\forall\) y
=> \(\left(x-1\right)^2+\left|y+2\right|\ge0\)\(\forall\)x,y
=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\y+2=0\end{cases}}\)
=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy ...
b) Ta có: \(\frac{1}{2}-\frac{y}{3}=\frac{2}{x}\)
=> \(\frac{3-2y}{6}=\frac{2}{x}\)
=> \(x\left(3-2y\right)=12\)
=> x; 3 - 2y \(\in\)Ư(12) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 12; -12}
Do 3 - 2y là số lẽ , mà x,y \(\in\)Z
=> 3 - 2y \(\in\) {1; -1; 3; -3}
Lập bảng :
| 3 - 2y | 1 | -1 | 3 | -3 |
| x | 12 | -12 | 4 | -4 |
| y | 1 | 2 | 0 | 3 |
Vậy ...
a, \(\frac{2}{5}+\frac{1}{4}\times x=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{3}{10}-\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}\times x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{1}{4}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
Vậy \(x=\frac{-2}{5}\)
b, \(\frac{2}{3}+\frac{2}{3}\div x=\frac{4}{15}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)
\(\Leftrightarrow\frac{2}{3}\div x=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{2}{3}\div\frac{-2}{5}\)
\(\Leftrightarrow\frac{-5}{3}\)
Vậy \(x=\frac{-5}{3}\)
c, \(2\times\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)
\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{2}\div2\)
\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{4}\\\frac{2}{3}-x=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{11}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{12};\frac{11}{12}\right\}\)
d, \(3\times\left|\frac{5}{4}-x\right|-\frac{1}{8}=\frac{1}{4}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{1}{4}+\frac{1}{8}\)
\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{3}{8}\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{3}{8}\div3\)
\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{1}{8}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-x=\frac{1}{8}\\\frac{5}{4}-x=\frac{-1}{8}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{9}{8}\\x=\frac{11}{8}\end{cases}}\)
Vậy \(x\in\left\{\frac{9}{8};\frac{11}{8}\right\}\)
a) \(\Leftrightarrow x+\frac{3}{4}x=\frac{1}{3}+\frac{5}{4}\)
\(\Leftrightarrow\frac{7}{4}x=\frac{19}{12}\Leftrightarrow x=\frac{19}{12}:\frac{7}{4}=\frac{19}{21}\)
b) \(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x=\frac{1}{4}+\frac{1}{5}\Leftrightarrow\frac{1}{6}x=\frac{9}{20}\Leftrightarrow x=\frac{9}{20}:\frac{1}{6}=\frac{27}{10}\)