\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{7}{5}\)

b) \(-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)\)

\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{41}{15}\)

\(\Rightarrow x=-\dfrac{41}{10}\)

c) \(\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5\)

\(\Leftrightarrow\dfrac{8+15x}{10x}.\dfrac{3}{2}=\dfrac{-3}{2}\)

\(\Leftrightarrow\dfrac{24+45x}{20x}=\dfrac{-3}{2}\)

\(\Leftrightarrow-60x=48+90x\)

\(\Rightarrow x=-0,32\)

d) \(\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x\)

\(\Leftrightarrow\dfrac{4x-2}{3}=\dfrac{1-4x}{4}\)

\(\Rightarrow16x-8=3-12x\)

\(\Rightarrow x=\dfrac{11}{28}\)

a: \(\dfrac{2.75}{x}=\dfrac{0.4}{1.5}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{11}{4}\cdot\dfrac{15}{4}=\dfrac{165}{16}\)

b: \(3\dfrac{1}{2}:\left(2x-3\right)=\dfrac{-3}{4}:0.2\)

\(\Leftrightarrow\dfrac{7}{2}:\left(2x-3\right)=\dfrac{-3}{4}:\dfrac{1}{5}=\dfrac{-15}{4}\)

\(\Leftrightarrow2x-3=\dfrac{7}{2}:\dfrac{-15}{4}=\dfrac{-7}{2}\cdot\dfrac{4}{15}=\dfrac{-28}{30}=\dfrac{-14}{15}\)

=>2x=-14/15+3=45/45-14/15=31/45

=>x=31/90

c: \(\dfrac{3x+2}{27}=\dfrac{3}{3x+2}\)

\(\Leftrightarrow\left(3x+2\right)^2=81\)

=>3x+2=9 hoặc 3x+2=-9

=>3x=7 hoặc 3x=-11

=>x=7/3 hoặc x=-11/3

d: \(\dfrac{5-x}{4}=\dfrac{2x+3}{2}\)

=>10-2x=8x+12

=>-10x=2

hay x=-1/5

a.\(\left(3x-2\right)^2=16\)

Ta có: \(\left(3x-2\right)^2=16\)

\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)

\(\Rightarrow3x-2=4\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)

\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)

\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)

\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)

\(\Rightarrow x=\dfrac{7}{16}\)

c. \(5^{x+2}+5^x=3250\)

\(5^x.5^2+5^x=3250\)

\(5^x.\left(5^2+1\right)=3250\)

\(5^x.26=3250\)

\(5^x=125\)

\(5^x=5^3\)

\(x=5\)

TA CÓ:

\(\frac{3}{x-5}=\frac{4}{x+2}=\frac{3-\left(-4\right)}{x-5-x-2}=\frac{7}{-7}=-1\)

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

=>\(\frac{4}{x+2}=-1\)=>4=-x-2=>x=-6

Vì ko thể có 2 giá trị x trong 1 trường hợp nên ko tồn tại x thỏa mãn đề bài

a) \(\left(x+1\right)\left(x+3\right)-x\left(x+2\right)=7\)

\(\Leftrightarrow x^2+4x+3-x^2-2x=0\)

\(\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)

\(2x\left(3x+5\right)-x\left(6x-1\right)=33\)

\(\Leftrightarrow6x^2+10x-6x^2+x=33\)

\(\Leftrightarrow11x=33\Leftrightarrow x=3\)

\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)

\(\Leftrightarrow2x-1=\frac{5}{2}\)

\(\Leftrightarrow2x=\frac{7}{2}\)

\(\Leftrightarrow x=\frac{7}{4}\)