a.

\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

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

TH2:

\(x-\frac{3}{4}=0\)

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

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

b.

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

TH1:

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

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

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

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

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

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

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

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

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

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

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

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

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

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

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

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

\(x=5\)

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

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

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

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

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

=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)

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

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

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

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

\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)

\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)

\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)

\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)

\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)

\(A=1,3\left(2\right)\)

 a)     \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)

<=>\(\hept{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{4}\end{cases}}}\)

b)        \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}}\)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

â)\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(\Leftrightarrow27x-18=2x-3x^2\)

\(\Leftrightarrow27x-18-2x+3x^2=0\)

\(\Leftrightarrow3x^2+25x-18=0\)

\(\Leftrightarrow3x^2+27x-2x-18=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)

       \(\Rightarrow\orbr{\begin{cases}3x-2=0\\x+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)

b)\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)

\(\Leftrightarrow4x^2-4x+1-4x^2+25=18\)

\(\Leftrightarrow26-4x=18\)

 \(\Leftrightarrow4x=8\)

      \(\Rightarrow x=2\)

c)\(5x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow5x^2-10x-2x+10=0\)

\(\Leftrightarrow5x^2-12x+10=0\)

\(\Leftrightarrow x^2-6x+2=0\)

\(\Leftrightarrow x^2-6x+9-7=0\)

\(\Leftrightarrow\left(x-3\right)^2=7\)

       \(\Rightarrow\orbr{\begin{cases}x-3=\sqrt{7}\\x-3=-\sqrt{7}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}+3\\x=-\sqrt{7}+3\end{cases}}\)

d)\(x^2-5=0\)

\(\Leftrightarrow x^2=5\)

    \(\Rightarrow x=\sqrt{5};-\sqrt{5}\)

e)\(x^3+5x^2-4x-20=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+10x-20=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+10\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x-2\right)=0\)

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

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

      \(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=-2;2\end{cases}}\)

a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)

Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)

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

\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)

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

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

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

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

\(\Rightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)

\(Ta\) \(có\)

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

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

\(3x+2=4x\)

\(3x+2=3x+x\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) Vì \(\left|x+\frac{1}{5}\right|\ge0;\left|x+\frac{2}{5}\right|\ge0;\left|x+1\frac{2}{5}\right|\ge0\forall x\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

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

\(\Rightarrow3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)

\(\Rightarrow2=4x-3x=x\)

Vậy x = 2

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

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

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

\(\Rightarrow\frac{4}{6}-\frac{2x}{6}+\frac{3}{6}-\frac{6x}{6}-\frac{3}{6}=\frac{30}{6}\)

\(\Rightarrow4-2x+3-6x-3=30\)

\(\Rightarrow4-8x=30\)

\(\Rightarrow-8x=26\)

\(\Rightarrow x=\frac{26}{-8}=-\frac{13}{4}\)

Vậy \(x=-\frac{13}{4}\)