(3,5 + 2x).2 2/3 = 5 1/3 => (3,5 + 2x).8/3 = 16/3 => 3,5 + 2x = 16/3 : 8/3 = 2/3 => 21/6 + 2x = 4/6 => 2x = 4/6 - 21/6 = -17/6 => x = -17/6 : 2 = -17/12

6/10 và 5/10

56/126 và 81/126

15/30 và 20/30 và 18/30

\(\dfrac{1}{2};\dfrac{2}{3};\dfrac{3}{5}\left(MSC:30\right)\)

\(\dfrac{1}{2}=\dfrac{1\times15}{2\times15}=\dfrac{15}{30}\)

\(\dfrac{2}{3}=\dfrac{2\times10}{3\times10}=\dfrac{20}{30}\)

\(\dfrac{3}{5}=\dfrac{3\times6}{5\times6}=\dfrac{18}{30}\)

6/10 và 5/10.

56/126 và 81/126.

15/30 và 20/30 và 18/30.

[3,8 : (x + 1,05)] : \(\frac{1}{3}\)= 2\(\frac{1}{9}\)

\(\left[\frac{19}{5}:\left(x+\frac{21}{20}\right)\right]:\frac{1}{3}=\frac{19}{9}\)

\(\frac{19}{5}:\left(x+\frac{21}{20}\right)=\frac{19}{9}.\frac{1}{3}=\frac{19}{27}\)

\(x+\frac{21}{20}=\frac{19}{5}:\frac{19}{27}=\frac{27}{5}\)

\(x=\frac{27}{5}-\frac{21}{20}\)

\(x=\frac{87}{20}\)

a,\(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\) 

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

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

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

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

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

\(\Rightarrow\frac{x+3}{x-2}=0\Leftrightarrow x+3=0\Rightarrow x=-3\left(tm\right)\)

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

\(\Rightarrow\left(x-3\right)\left(2x+1\right)-\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x-3\right)\left(2x+1-x+1\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

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

c, \(|x-5|=2x+1\)

\(\Rightarrow|x-5|=\orbr{\begin{cases}x-5\Leftrightarrow x-5\ge0\Rightarrow x\ge5\\x-5\Leftrightarrow x-5< 0\Rightarrow x< 5\end{cases}}\)

Nếu \(x\ge5\Rightarrow x-5=2x+1\)

\(\Rightarrow x=-6\left(ktm\right)\)

Nếu \(x< 5\Rightarrow-\left(x-5\right)=2x+1\)

\(\Rightarrow-x+5=2x+1\)

\(\Rightarrow3x=4\)

\(\Rightarrow x=\frac{4}{3}\left(tm\right)\)

Vậy \(S=\left\{\frac{4}{3}\right\}\)

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

=> \(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=> \(3x-\frac{1}{2}=2x+\frac{9}{2}\)

=> \(3x-\frac{1}{2}-2x-\frac{9}{2}=0\)

=> \(\left(3x-2x\right)+\left(-\frac{1}{2}-\frac{9}{2}\right)=0\)

=> x - 5 = 0 => x = 5

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

=> \(\frac{2}{3}-\frac{1}{3}\left[x\cdot\left(-\frac{2}{3}\right)\right]-x-\frac{1}{2}=5\)

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

=> \(\frac{2}{3}-\left(-\frac{2x}{9}\right)-x-\frac{1}{2}=5\)

=> \(\frac{2}{3}+\frac{2x}{9}-x-\frac{1}{2}=5\)

=> \(\frac{6}{9}+\frac{2x}{9}-x=\frac{11}{2}\)

=> \(\frac{6+2x}{9}-x=\frac{11}{2}\)

=> \(\frac{6+2x}{9}-\frac{9x}{9}=\frac{11}{2}\)

=> \(\frac{6-7x}{9}=\frac{11}{2}\)

=> \(6-7x=\frac{11}{2}\cdot9=\frac{99}{2}\)

=> \(7x=6-\frac{99}{2}\)

=> \(7x=-\frac{87}{2}\)

=> \(x=\left(-\frac{87}{2}\right):7=\left(-\frac{87}{2}\right)\cdot\frac{1}{7}=-\frac{87}{14}\)

p/s : nếu gõ kiểu latex \(\frac{x}{-\frac{3}{2}}\)thì cái đây không phải là hỗn số đâu.Bạn nên kiểm tra đề :)))

a) (x-1)*(x+2)-(x-3)*(-x+4)=19

\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)

\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)

\(\Leftrightarrow2x^2-3x+7-19=0\)

\(\Leftrightarrow2x^2-3x-12=0\)

Đề sai??

b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)

\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)

\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)

\(\Leftrightarrow-30x^2+7x-4+17=0\)

\(\Leftrightarrow-30x^2+7x+13=0\)

???

c) (x+1)*(x+1)-(x-1)*(x-1)=9

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=9\)

\(\Leftrightarrow\left(x+1+x-1\right)\left(x+1-x+1\right)=9\)

\(\Leftrightarrow2x.2=9\)

\(\Leftrightarrow x=\frac{9}{4}\)

A, \(2\frac{2}{5}\left(\frac{1}{2}x-0,75\right)=\frac{3}{10}\)

\(=>\frac{2.5+2}{5}\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(=>\frac{1}{2}x-\frac{3}{4}=\frac{3}{10}:\frac{12}{5}=\frac{1}{8}\)

\(=>x=\left(\frac{1}{8}+\frac{3}{4}\right):\frac{1}{2}\)

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

B, \(\frac{3}{5}-|x-\frac{1}{2}|=25\%\)

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

\(=>|x-\frac{1}{2}|=\frac{7}{20}\)

\(=>x-\frac{1}{2}=\frac{7}{20};-\frac{7}{20}\)

TH1: \(x-\frac{1}{2}=\frac{7}{20}=>x=\frac{17}{20}\)

TH2: \(x-\frac{1}{2}=-\frac{7}{20}=>x=\frac{3}{20}\)