giúp với

c) \(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)

Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm

\(a,3^4:3^x=3^3\)

\(3^{4-x}=3^3\)

=> 4 -x = 3

=> x = 1

Nguyễn Tấn Tài bn làm hết nhé

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

\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)

\(\Rightarrow2x-\frac{1}{2}=\pm\left(\frac{13}{35}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{13}{35}\\2x-\frac{1}{2}=\frac{-13}{35}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{61}{70}\\2x=\frac{9}{70}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{61}{140}\\x=\frac{9}{140}\end{cases}}\)

~Study well~

#KSJ

\(b,\frac{3}{4}-4\times\left|2x+1\right|=\frac{1}{2}\)

\(\Rightarrow4\times\left|2x+1\right|=\frac{1}{4}\)

\(\Rightarrow\left|2x+1\right|=\frac{1}{16}\)

\(\Rightarrow2x+1=\pm\left(\frac{1}{16}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{16}\\2x+1=\frac{-1}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{-15}{16}\\2x=\frac{-17}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{32}\\x=\frac{-17}{32}\end{cases}}\)

~Study well~

#KSJ

K chép lại đề, lm luôn nhé:

*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)

\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)

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

* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)

=> K có gt x nào t/m đề

* Đề sai

* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)

\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)

\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)

* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)

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

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

\(\Rightarrow-7x=-\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)

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

\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)

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

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

\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)

\(\Rightarrow x\in\varnothing\)

c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.

d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)

\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)

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

\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)

\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)

\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)

g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)

\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)

\(th1:x=0\)

\(th2:x=\dfrac{-3}{5}\)

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

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

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

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

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

\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)

a) Ta có:

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

\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)

\(\Rightarrow6x+3=5x+10\)

\(\Rightarrow6x-5x=10-3\)

\(\Rightarrow x=7\)

b)Ta có:

\(\frac{5}{8x-2}=\frac{-4}{7-x}\)

\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)

\(\Rightarrow35-5x=-32x+8\)

\(\Rightarrow-5x+32x=8-35\)

\(\Rightarrow27x=-27\)

\(\Rightarrow x=-1\)

c) Ta có:

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

\(\Rightarrow4x=3\left(2x-1\right)\)

\(\Rightarrow4x=6x-3\)

\(\Rightarrow3=6x-4x=2x\)

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

d)Ta có:

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

\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)

\(\Rightarrow8x-4=9x+3\)

\(\Rightarrow8x-9x=3+4\)

\(\Rightarrow-x=7\Rightarrow x=-7\)

e)Ta có:

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

\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)

\(\Rightarrow12x+4=7x+14\)

\(\Rightarrow12x-7x=14-4\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

f)Ta có:

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

\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)

\(\Rightarrow-6+6x=4x+4\)

\(\Rightarrow6x-4x=4+6\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

\(\left(2x-3\right)\left(6-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=3\end{matrix}\right.\)

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

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

\(\Leftrightarrow\frac{1}{3\left(2x-1\right)}=\frac{-21}{4}\)

\(\Leftrightarrow-63\left(2x-1\right)=4\)

\(\Leftrightarrow2x-1=-\frac{4}{63}\)

\(\Leftrightarrow2x=\frac{59}{63}\)

\(x=\frac{59}{126}\)

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

\(\Rightarrow2x=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)

\(\Rightarrow x=\frac{5}{4}:2=\frac{5}{8}\)

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

\(\Rightarrow x-5=\frac{2}{5}+\frac{1}{3}=\frac{11}{15}\)

\(\Rightarrow x=\frac{11}{15}+5=5\frac{11}{15}\)

\(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)

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

\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)

|2x-5|=4

 Suy ra: 2x-5=4

          hoặc 2x-5 =-4

nếu 2x-5=4                                                     nếu 2x - 5=-4

2x=4+5                                                               2x=(-4)+5

2x=9                                                                   2x=-1

x=9:2                                                                  x=(-1):2

x=4/5(loại)                                                          x=-1/2 (loại)