( 2x - 3 )( x + 1 ) - 2x² + 6x = 0

<=> 2x² - x - 3 - 2x² + 6x = 0

<=> 5x - 3 = 0

<=> 5x = 3

<=> x = 3/5

( x² - x + 1 )( x - 3 ) - x³ + 4x² = 0

<=> x³ - 4x² + 4x - 3 - x³ + 4x² = 0

<=> 4x - 3 = 0

<=> 4x = 3

<=> x = 3/4

( x² - 2 )( x² + 2 ) - x⁴ - 2x + 5 = 0

<=> ( x² )² - 4 - x⁴ - 2x + 5 = 0

<=> x⁴ + 1 - x⁴ - 2x = 0

<=> 1 - 2x = 0

<=> 2x = 1

<=> x = 1/2

( x - 3 )( x² - 3x + 2 ) - ( x² - 2x - 7 )( x - 2 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - ( x³ - 4x² - 3x + 14 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - x³ + 4x² + 3x - 14 + 2x² - 2x = 0

<=> 12x - 20 = 0

<=> 12x = 20

<=> x = 20/12 = 5/3

a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)

\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)

b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)

\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)

c ; d tương tự nhé ! 

chúc bạn học giỏi

a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)

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

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

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

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

hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)

b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)

\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)

=>6=0(vô lý)

c: \(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)

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

=>x=-2 hoặc x=2

đ: \(\Rightarrow2x^2-2x-5x+5=0\)

=>(x-1)(2x-5)=0

=>x=1 hoặc x=5/2

Trần văn ổi ()

đù khó thế

tl j z mấy chế , k câu dc đâu :))

sửa lại chút: a) (2x+1)^2-2x-1=2              b) (x^2-3x)^2+5(x^2-3x)+6=0                  c) (x^2-x-1)(x^2-x)-2=0            d) (5-2x)^2+4x-10=8                   e) (x^2+2x+3)(x^2+2x+1)=3              f) x(x-1)(x^2-x+1)-6=0

a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)

a, (\(x-2\))² - (2\(x\) + 3)² = 0

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

     (-\(x\) - 5)(3\(x\) +1) = 0

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

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

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

Vậy \(x\in\) { -5;- \(\dfrac{1}{3}\)}

b, 9.(2\(x\) + 1)² - 4.(\(x\) + 1)² = 0 

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

    (6\(x\) + 3 - 2\(x\) - 2)(6\(x\) + 3 + 2\(x\) + 2) = 0

      (4\(x\) + 1)(8\(x\) + 5) =0

        \(\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{8}\end{matrix}\right.\)

          S = { - \(\dfrac{5}{8}\); \(\dfrac{-1}{4}\)}

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

      \(x\left(x+1\right)\).(\(x\) - 1) + \(x\)(\(x\) -1) = 0

        \(x\)(\(x\) -1)(\(x\) + 1 + 1) = 0

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

             \(\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\)

               \(\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)

              S = { -2; 0; 1}

e, (\(x\) - 2)²- (\(x\) - 2)(\(x\) + 2) = 0 

     (\(x\) - 2)(\(x-2\) - \(x\) - 2) =0

      -4 (\(x-2\)) = 0

            \(x\) - 2 = 0

            \(x\) = 2

          S ={ 2}

1)\(2x+6=0\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy : x=3 là nghiệm PT

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

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

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

Vậy:....

3)\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

\(\Rightarrow\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

\(\Leftrightarrow x^2-4x+4+3x+6-x^2+11=0\)

\(\Leftrightarrow-x+21=0\)

\(\Leftrightarrow-x=-21\)

\(\Leftrightarrow x=21\)

Vậy:......

4) \(x\left(x^2-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x^2=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy:........

5)\(4x+20=0\)

\(\Leftrightarrow4x=-20\)

\(\Leftrightarrow x=-5\)

Vậy:...

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

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

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

\(\Leftrightarrow-2=0\)(vô lí)

Vậy : PT vô nghiệm

7)\(\frac{1+2x-5}{6}=\frac{3-x}{4}\)

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

\(\Rightarrow2\left(-4+2x\right)=3\left(3-x\right)\)

\(\Leftrightarrow-8+4x-9+3x=0\)

\(\Leftrightarrow-17+7x=0\)

\(\Leftrightarrow7x=17\)

\(\Leftrightarrow x=\frac{17}{7}\)

8) Làm tương tự

9) \(2\left(x+1\right)=5x-7\)

\(\Leftrightarrow2x+2-5x+7=0\)

\(\Leftrightarrow-3x+9=0\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\)

#H

1.\(2x+6=0\)

\(\Leftrightarrow2\left(x+3\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)

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

\(\Leftrightarrow\left(x-1\right)^2-4=0\)

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

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

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

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

Vậy tập nghiệm của PT là \(S=\left\{3;-1\right\}\)

3.\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

ĐKXĐ :\(x\ne\pm2\)

Ta có ; \(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

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

\(\Leftrightarrow\frac{x^2-4x+4+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x^2-x+10}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-x+10=x^2-11\)

\(\Leftrightarrow21-x=0\)

\(\Leftrightarrow x=21\)(Thỏa mãn ĐKXĐ)

Vậy tập nghiệm của PT là \(S=\left\{21\right\}\)

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

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

\(\Leftrightarrow x=0\)

hoặc \(x-1=0\)

hoặc \(x+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Vậy tập nghiệm của PT là \(S=\left\{0;\pm1\right\}\)

5.\(4x+20=0\)

\(\Leftrightarrow4\left(x+5\right)=0\)

\(\Leftrightarrow x+5=0\)

\(\Leftrightarrow x=-5\)

Vậy tập nghiệm của PT là \(S=\left\{-5\right\}\)

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

ĐKXĐ : \(x\notin\left\{-1;0\right\}\)

Ta có : \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)

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

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

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

\(\Rightarrow2x^2+2x-2=2x^2+2x\)

\(\Leftrightarrow0x=2\)(Vô lí)

Vậy PT vô nghiệm 

7.\(1+\frac{2x-5}{6}=\frac{3-x}{4}\)

\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}=\frac{3\left(3-x\right)}{12}\)

\(\Leftrightarrow\frac{12+4x-10}{12}=\frac{9-3x}{12}\)

\(\Leftrightarrow\frac{4x+2}{12}=\frac{9-3x}{12}\)

\(\Rightarrow4x+2=9-3x\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

Vậy tập nghiệm của PT là \(S=\left\{1\right\}\)

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

ĐKXĐ : \(x\notin\left\{0;2\right\}\)

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

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

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

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

\(\Rightarrow x^2+x+2=2\)

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

\(\Leftrightarrow x\left(x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)(Không thỏa mãn ĐKXĐ)_(Thỏa mãn ĐKXĐ)

Vậy tập nghiệm của PT là \(S=\left\{-1\right\}\)

9.\(2\left(x+1\right)=5x-7\)

\(\Leftrightarrow2x+2=5x-7\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)