\(A.\)\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
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18 tháng 6 2017

A . \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

                    \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

                                \(x=\frac{1}{4}-\frac{2}{5}=\frac{-3}{20}\)

B . \(2.x.\left(x-\frac{1}{7}\right)=0\)

    \(\Rightarrow x.\left(x-\frac{1}{7}\right)=0\)

                        \(\Rightarrow x=0\)

C .\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

                \(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)

                 \(\frac{1}{4}:x=\frac{-7}{20}\)

                        \(x=\frac{1}{4}:\frac{-7}{20}\)

                         \(x=\frac{-5}{7}\)

18 tháng 6 2017

giúp mk nhanh đi mn

mk cần gấp lắm ak

Bài 1:

a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)

\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)

\(=\frac{2+3}{12}=\frac{5}{12}\)

b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)

\(=\frac{23}{12}-\frac{-7}{10}\)

\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)

Bài 2:

a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)

\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)

\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)

\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)

\(\Leftrightarrow-2x=-6-1=-7\)

hay \(x=\frac{7}{2}\)

Vậy: \(x=\frac{7}{2}\)

14 tháng 8 2020

lớp 9 đấy!

28 tháng 4 2019

4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)

\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)

Vậy x=\(\frac{3}{-20}\)

b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)

\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)

Vậy x=\(\frac{1}{-2}\)

g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)

\(\Leftrightarrow\left|4x-1\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)

i) \(\left(x-1^3\right)=125\)

\(\Leftrightarrow x-1=125\)

\(\Leftrightarrow x=125+1=126\)

Vậy x=126

k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

29 tháng 4 2019

xin lỗi câu h tui xin chữa lại là:\(|x+70\%|=2\frac{1}{5}\)

24 tháng 6 2020

bạn tự làm đi tính toán thôi mà

19 tháng 4 2019

Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

19 tháng 4 2019

g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

9 tháng 3

\(a;\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}=\dfrac{20}{27}\\ b;\left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(-\dfrac{10}{11}\right)\\ \dfrac{9}{11}-x=-\dfrac{2}{11}\\ x=\dfrac{9}{11}-\left(-\dfrac{2}{11}\right)=\dfrac{9}{11}+\dfrac{2}{11}\\ x=1\\ c;-\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\\ -\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\\ -\dfrac{11}{12}x=-\dfrac{11}{12}\\ x=\left(-\dfrac{11}{12}\right):\left(-\dfrac{11}{12}\right)=1\)

\(d;-\dfrac{5}{4}-\left(1\dfrac{1}{2}+x\right)=4,5\\ \Leftrightarrow-\dfrac{5}{4}-\left(\dfrac{3}{2}+x\right)=4,5\\\dfrac{3}{2}+x=-\dfrac{5}{4}-4,5\\ \dfrac{3}{2}+x=-\dfrac{23}{4}\\ x=-\dfrac{23}{4}-\dfrac{3}{2}\\ x=-\dfrac{29}{4}\\ đ;\left(\dfrac{3}{4}-x:\dfrac{2}{15}\right)\cdot\dfrac{1}{5}=-2,6\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-2,6:\dfrac{1}{5}\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-13\\ x:\dfrac{2}{15}=\dfrac{3}{4}-\left(-13\right)\\ x:\dfrac{2}{15}=\dfrac{55}{4}\\ x=\dfrac{55}{4}\cdot\dfrac{2}{15}=\dfrac{11}{6}\\ e;3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}=-\dfrac{10}{3}\)

\(f;\left(1-2x\right)\cdot\dfrac{4}{5}=\left(-2\right)^3\\ \left(1-2x\right)\cdot\dfrac{4}{5}=-8\\ 1-2x=-8:\dfrac{4}{5}=-10\\ 2x=1-\left(-10\right)=11\\ x=\dfrac{11}{2}\\ g;\dfrac{1}{6}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{24}\Rightarrow x=\dfrac{3}{4}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{24}\Rightarrow x=\dfrac{7}{12}\end{matrix}\right.\)