a.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

b.\(\dfrac{2}{3}-\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

Đề là vậy đk bạn?

b.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{3}{2x-30}\right)=11\);\(x\ne15\)

\(\Leftrightarrow-\dfrac{3}{2x-30}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21\left(x-15\right)}{2\left(x-15\right)}\)

\(\Leftrightarrow-3=21\left(x-15\right)\)

\(\Leftrightarrow-3=21x-315\)

\(\Leftrightarrow21x=312\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

c.\(\dfrac{2}{3}-\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow-\left|4x+\dfrac{1}{2}\right|=-\dfrac{1}{6}\)

\(\Leftrightarrow\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{1}{2}=\dfrac{1}{6}\\4x+\dfrac{1}{2}=-\dfrac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

mn giúp mik với

ai nhanh nhất mik cho 5 sao

c) x^2 -x-20=0

\(\Leftrightarrow x^2-5x+4x-20=0\)

\(\Leftrightarrow\left(x^2+4x\right)-\left(5x+20\right)=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

Vậy...

Dễ thế mà không làm được

$\left|\frac54x-\frac72\right|=\frac58+\frac35$

$\left|\frac54x-\frac72\right|=\frac{49}{40}$

$\frac54x-\frac72=\pm\frac{49}{40}$

TH1:

$\frac54x-\frac72=\frac{49}{40}$

$\frac54x=\frac{189}{40}$

$x=\frac{189}{50}$

TH2:

$\frac54x-\frac72=-\frac{49}{40}$

$\frac54x=\frac{91}{40}$

$x=\frac{91}{50}$

Vậy $\boxed{x=\frac{91}{50};\frac{189}{50}}$.

a) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|\frac{5}{4}x-\frac{7}{2}\right|=0\\\left|\frac{5}{8}x+\frac{3}{5}\right|=0\end{cases}}=\hept{\begin{cases}\left|\frac{5}{4}x\right|=\frac{7}{2}\\\left|\frac{5}{8}x\right|=\frac{-3}{5}\end{cases}=\hept{\begin{cases}x=\frac{14}{5}\\x=\frac{-24}{25}\end{cases}}}\)

b) \(\left|\frac{7}{8}x+\frac{5}{6}\right|-\left|\frac{1}{2}x+5\right|=0\) 

\(\Rightarrow\hept{\begin{cases}\left|\frac{7}{8}x+\frac{5}{6}\right|=0\\\left|\frac{1}{2}x+5\right|=0\end{cases}}=\hept{\begin{cases}\left|\frac{7}{8}x\right|=\frac{-5}{6}\\\left|\frac{1}{2}x\right|=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-20}{21}\\x=\frac{-5}{2}\end{cases}}\)