a: \(\Leftrightarrow\left(x-\dfrac{2}{5}\right):\dfrac{3}{2}=-\dfrac{5}{4}+\dfrac{5}{2}=\dfrac{5}{4}\)

\(\Leftrightarrow x-\dfrac{2}{5}=\dfrac{5}{4}\cdot\dfrac{3}{2}=\dfrac{15}{8}\)

hay x=91/40

b: \(\Leftrightarrow\left(2.5x-3.6\right)=-1\cdot\dfrac{12}{7}=\dfrac{-12}{7}\)

=>2,5x=66/35

hay x=132/175

c: \(\Leftrightarrow\left(\dfrac{15}{4}-2x\right)=\dfrac{19}{9}:\dfrac{4}{3}=\dfrac{19}{9}\cdot\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x=15/4-19/12=45/12-19/12=26/12

=>x=13/12

\(\left(x-2,5\right)^2=\frac{4}{9}\)

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

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

\(x=\frac{2}{3}+2,5\)

\(x=\frac{19}{6}\)

\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)

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

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

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

\(2x=-1\)

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

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

\(\Rightarrow\)\(\orbr{\begin{cases}x-2,5=\frac{2}{3}\\x-2,5=-\frac{2}{3}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}+\frac{5}{2}\\x=-\frac{2}{3}+\frac{5}{2}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{4}{6}+\frac{15}{6}=\frac{19}{6}\\x=-\frac{4}{6}+\frac{15}{6}=\frac{11}{6}\end{cases}}\)

Ta có:

a)\(\left(x-2,5\right)^2=\frac{4}{9}\)

    \(x-2,5=\sqrt{\frac{4}{9}}\)

   \(x-2,5=\frac{2}{3}\)

   \(x=\frac{2}{3}+2,5=\frac{19}{6}\)

b) \(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)

    \(2x+\frac{1}{3}=\sqrt[3]{\frac{-8}{27}}\)

   \(2x+\frac{1}{3}=\frac{-2}{3}\)

   \(x=\left(\frac{-2}{3}-\frac{1}{3}\right)\div2=\frac{-1}{2}\)

\(x\div4\frac{1}{3}=-2,5\)

\(\Leftrightarrow x\div\frac{13}{3}=\frac{-5}{2}\)

\(\Leftrightarrow x=\frac{-5}{2}.\frac{13}{3}\)

\(\Leftrightarrow x=\frac{-65}{6}\)

\(x\div\frac{-3}{5}=\frac{-10}{21}\)

\(\Leftrightarrow x=\frac{-10}{21}.\frac{-3}{5}\)

\(\Leftrightarrow x=\frac{30}{105}\)

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

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

\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}+\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{6}{10}\)

\(\Leftrightarrow x=\frac{6}{10}\div\frac{2}{3}\)

\(\Leftrightarrow x=\frac{18}{20}\)

\(\Leftrightarrow x=\frac{9}{10}\)

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

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(x+1\right)=\frac{5}{2}\div\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)=5\)

\(\Leftrightarrow x=5-1\)

\(\Leftrightarrow x=4\)

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

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

\(\Leftrightarrow1-\frac{2x}{3}=\frac{3}{2}\)

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

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

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

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

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

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

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

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

\(\Leftrightarrow x=\frac{-25}{8}\)

1) \(|5x-3|=|7-x|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

Vậy...

2) \(2.|3x-1|-3x=7\)

\(\Leftrightarrow2.|3x-1|=7+3x\)

\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)

Vậy...

5) \(\text{1,6 - |x + 1,5| = 0}\)

\(|x + 1,5| = 1,6-0\)

\(\text{|x + 1,5| = 1,6}\)

\( |x | = 1,6-1,5\)

\(|x|=0,1\)

a: Ta có: \(8x+11-3=5x+x-3\)

\(\Leftrightarrow8x+8=6x-3\)

\(\Leftrightarrow2x=-11\)

hay \(x=-\dfrac{11}{2}\)

b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)

\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)

\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)

\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow-10x+2=0\)

\(\Leftrightarrow-10x=-2\)

hay \(x=\dfrac{1}{5}\)

d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)

\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)

\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)

hay t=2

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

a, |-2,5 + x| = 1,3

=> -2,5 + x = 1,3 hoặc -2,5 + x = -1,3

=> x = 3,8 hoặc x = 1,2

vậy_

b,