\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)

Vậy .........

\(b,\left(x^2-4\right)+\left(x-2\right)\left(3-2x=0\right)\)

\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)

Vậy ..................

\(c,x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

\(d,x\left(2x-7\right)-4x+14=0\)

\(\Leftrightarrow2x^2-7x-4x+14=0\)

\(\Leftrightarrow2x^2-11x+14=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)

Vậy ............

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

\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)

\(\Leftrightarrow3x^2-24x+21=0\)

\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy .....................

\(f,x^2-x-\left(3x-3\right)=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy ..............

a, 8/x-8 + 11/x-11 = 9/x-9  + 10/ x-10

b, x/x-3 - x/x-5 = x/x-4 - x/x-6

c, 4/x^2-3x+2  - 3/2x^2-6x+1   +1 = 0

d, 1/x-1 + 2/ x-2  + 3/x-3  = 6/x-6

e, 2/2x+1 - 3/2x-1 = 4/4x^2-1

f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3

g, 1/x-1 + 2x^2 -5/x^3 -1  = 4/ x^2 +x+1

\(a,\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

Vậy ............

\(b,\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2,3x=6,9\\0,1x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

Vậy ...........

\(c,\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-0,5\\x\in\varnothing\end{cases}}\)

Vậy .........................

\(d,\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+7=0\\x-5=0\\5x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-7\\x=5\\5x=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{cases}}\)

Vậy ...............

a) \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\)

\(\Leftrightarrow x=-\frac{1}{2}\) ( do \(x^2+1\ge1>0\forall x\) )

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-5=0\end{cases}hoặc5x+1=0}\)

\(\Leftrightarrow x\in\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}\)

a) \(7+2x=22-3x\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)

b) \(8x-3=5x+12\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

c) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow3x=36\Leftrightarrow x=12\)

d) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow3x=24\Leftrightarrow x=8\)

e) \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow x=7\)

f) \(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow0x=9\) ( vô lí )

8x² - 4x = 0 

=> 4x ( 2x - 1 ) = 0

<=>\(\orbr{\begin{cases}4x=0\\2x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

KL. Tập Ngiệm của pt ............

tự làm nốt 

8x²-4x=0

=>4x.2x-4x=0

=>4x.(2x-1)=0

=>4x=0 hoặc 2x-1 =0

   x    =0:4      2x     =0+1

 x      =0          2x    =1

                       x       =1:2=0,5

Vậy x\(\in\){0;0,5}

b)-6x+9x²=0

  -3x.2+3x.3x.3=0

=>3x.(-2+1.3x+3)=0

=>3x=0 hoặc -2+1.3x+3=0

      x  =0:3            3x+3 =0+2

   x    =0                3x+3  =2

                             3x       =2-3

                            3x      =-1

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

Vậy x\(\in\){0;\(\frac{-1}{3}\)}

4x²=3x

=>4x²-3x=0

=>x.(4x-3)=0

=>x=0 hoặc 4x-3=0

    x=0          4x    =0+3

                   4x     =3

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

Vậy x\(\in\){0;\(\frac{3}{4}\)}

Các phần khác bạn làm tương tự nha

Chúc bn học tốt

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

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

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

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

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

\(\Leftrightarrow-x^3-x^2+3x+1=-x^3+x^2+x-1\)

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

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{cases}}\)

Vay...

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-4x+4-3x-6=2x-22\)

\(\Leftrightarrow x^2-7x-2-2x+22=0\)

\(\Leftrightarrow x^2-9x+20=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

\(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)

\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)

\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)

\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)

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

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

b) Thay x = -3 vào A, ta được :

\(A=\frac{\left(-2\right)\left(-3\right)\left(-3-1\right)}{\left(-3+1\right)^2}\)

\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)

\(\Leftrightarrow A=-6\)

c) Để A > -1

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

\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)

\(\Leftrightarrow2x^2-2x< x^2+2x+1\)

\(\Leftrightarrow x^2-4x-1< 0\)

\(\Leftrightarrow\left(x-2\right)^2-5< 0\)

\(\Leftrightarrow\left(x-2\right)^2< 5\)

Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)