\(c.\left(1-2x\right)^2-\left(3x-2\right)^2=0\)

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

\(\left(-5x+3\right)\left(x-1\right)=0\)

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

\(d.\left(x-2\right)^2-\left(5-2x\right)^2=0\)

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

\(\left(3x-7\right)\left(-x+3\right)=0\)

\(\left[{}\begin{matrix}3x-7=0\\-x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

\(c,\Leftrightarrow1-4x+4x^2=9x^2-12x+4\\ \Leftrightarrow5x^2-8x+3=0\\ \Leftrightarrow\left(x-1\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2-5+2x\right)\left(x-2+5-2x\right)=0\\ \Leftrightarrow\left(3x-7\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

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

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

\(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{4}:2=\dfrac{7}{8}\)

\(\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}=\dfrac{29}{24}\)

\(x=\dfrac{29}{24}:\dfrac{1}{2}=\dfrac{29}{12}\)

Vậy : ...

d. \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)

\(-\dfrac{1}{2}x=\dfrac{1}{10}-\dfrac{4}{5}=-\dfrac{7}{10}\)

\(x=-\dfrac{7}{10}:\left(-\dfrac{1}{2}\right)=\dfrac{7}{5}\)

Vậy : ...

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

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}\)

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{6}{4}\)

\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\)

\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}:2\)

\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}.\dfrac{1}{2}\)

\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{8}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{21}{24}+\dfrac{8}{24}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{29}{24}\)

\(\Rightarrow x=\dfrac{29}{24}:\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{29}{24}.2\)

\(\Rightarrow x=\dfrac{29}{12}\)

d) \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{4}{5}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{10}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}:\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{7}{10}.2\)

\(\Rightarrow x=\dfrac{7}{5}\)

a) \(\Leftrightarrow x^2-x-x^2+2x=5\)
    \(\Leftrightarrow x=5\)
b) \(\Leftrightarrow4x\left(x^2-9\right)=0\)
    \(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0 \)
    \(\Leftrightarrow\)\(\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
    \(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy x = 0 , x = 3 hoặc x = -3

\(a,\Leftrightarrow x^2-x-x^2+2x=5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow4x\left(x^2-9\right)=0\\ \Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-3^2-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-81=0\\ \Leftrightarrow\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)\left(x^2-9x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\\left(x-\dfrac{9}{2}\right)^2+\dfrac{23}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: =2x^3-3x-5x^3-x^2-x^2

=-3x^3-2x^2-3x

b: =2(x^2+x-6)+x^2-4x+4+x^2+6x+9

=2x^2+2x-12+2x^2+2x+13

=4x^2+4x+1

d: =4x^2-9-x^2-10x-25-x^2-x+2

=2x^2-11x-32

a: \(x^2=2\)

=>\(x^2=\left(\sqrt{2}\right)^2\)

=>\(x=\pm\sqrt{2}\)

b: \(x^2=9\)

=>\(x^2=3^2\)

=>\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(\left(x-\sqrt{2}\right)^2=2\)

=>\(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=0\end{matrix}\right.\)

d: \(4x^2-1=0\)

=>\(4x^2=1\)

=>\(x^2=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

`c,(x-2)(2x-1)-(2x-3)(x-1)-2`

`=2x^2-x-4x+2-2x^2+2x+3x-3-2`

`=-3`

`->` Biểu thức không phụ thuộc vào biến `x`