d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

1 

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

= \(3\cdot25:\frac{5}{4}\)

= \(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) = 

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

a.

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

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

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)

\(-\frac{5}{6}\times x=\frac{5}{2}\)

\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)

\(x=-3\)

b.

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)

\(3x-\frac{37}{10}=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+\frac{37}{10}\)

\(3x=\frac{-95+37}{10}\)

\(3x=-\frac{58}{10}\)

\(3x=-\frac{29}{5}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{5}\times\frac{1}{3}\)

\(x=-\frac{29}{15}\)

c.

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21-16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}\times\frac{4}{3}\)

\(x=\frac{5}{6}\)

d.

\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)

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

\(-\frac{2}{3}\times x=\frac{3-2}{10}\)

\(-\frac{2}{3}\times x=\frac{1}{10}\)

\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)

\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)

\(x=-\frac{3}{20}\)

e.

\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x\right|=\frac{20+9}{12}\)

\(\left|x\right|=\frac{29}{12}\)

\(x=\pm\frac{29}{12}\)

Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

f.

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

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

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

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

\(2x-\frac{1}{3}=\pm\frac{1}{6}\)

                \(2x=\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{1+2}{6}\)

                \(2x=\frac{3}{6}\)

                \(2x=\frac{1}{2}\)

                  \(x=\frac{1}{2}\div2\)

                  \(x=\frac{1}{2}\times\frac{1}{2}\)

                  \(x=\frac{1}{4}\)

                \(2x=-\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{-1+2}{6}\)

                \(2x=\frac{1}{6}\)

                 \(x=\frac{1}{6}\div2\)

                 \(x=\frac{1}{6}\times\frac{1}{2}\)

                 \(x=\frac{1}{12}\)

Vậy x = 1/4 hoặc x = 1/12.

Chúc bạn học tốt

Sorry nha, mik chép lộn đềLàm lại câu a nha

a.

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

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

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)

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

\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)

\(x=-\frac{1}{2}\)

Chúc bạn học tốt

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

\(\frac{4}{6}x-\frac{9}{6}x=\frac{5}{12}\)

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

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

\(x=\frac{-1}{2}\)

\(b,\frac{2}{5}+\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{4}{10}\)

\(3x-3,7=\frac{-57}{10}:\frac{3}{5}\)

\(3x=\frac{-19}{2}+\frac{37}{10}\)

\(3x=\frac{-95}{10\text{​​}\text{​​}\text{​​}}+\frac{37}{10}\)

\(x=\frac{-29}{5}:3\)

\(x=\frac{-29}{15}\)

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}

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