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

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

\(3x-1=2\)

\(3x=3\)

\(x=1\)

1: \(\left(2x-2\right)\left(3x+1\right)-\left(3x-2\right)\left(2x-3\right)=5\)

=>\(6x^2+2x-6x-2-\left(6x^2-9x-4x+6\right)=5\)

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

=>\(3x-5-9x^2+15x+\left(2x-4\right)\left(3x-2\right)=x-6\)

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

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

=>\(-3x^2+x+9=0\)

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

3: \(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(2x+1\right)\left(4x^2-2x+1\right)=5x+6\)

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

Bài 1:

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

b) \(-\left(5+4y\right)\left(5-4y\right)=-\left[\left(5+4y\right)\left(5-4y\right)\right]=-\left[5^2-\left(4y\right)^2\right]=-\left(25-16y^2\right)=-25+16y^2\)

\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)

\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)

\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)

\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)

\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)

\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)

\(\Leftrightarrow4x^2+6x-51=0\)

\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

1 ) 3x^2 - 11x + 6 = 3x^2 - 9x - 2x + 6 = 3x( x- 3  ) - 2( x - 3) = ( 3x - 2 )( x - 3 )

2) 8x^2 - 2x - 1 = 8x^2 - 4x + 2x - 1 = 4x(  2x - 1 ) + 2x - 1 = ( 4x + 1 )( 2x - 1 )

3; 8x^2 - 2x - 1 =8x^2 - 4x + 2x - 1 = 4x(  2x - 1 ) + 2x - 1 = ( 4x + 1 )( 2x - 1 )

4; x^4 - 3x^2 - 4 = x^4 - 4x^2 + x^2 - 4 = x^2 ( x ^2 - 4 ) + x^2 - 4 = ( x^2 + 1 )( x^2 - 4 ) = ( x^2 + 1 )( x - 2 )( x + 2)

5) = x^2 ( x + 2 ) - 3 ( x+  2 ) = ( x^2 - 3 )( x + 2 ) 

Nhiều quá 

c. - x ( x + 3 ) + 2 = ( 4x + 1 ) ( x - 1 ) + 2x

<=> - x² - 3x + 2 = 4x² - x - 1

<=> 4x² - x - 1 + x² + 3x - 2 = 0

<=> 5x² + 2x - 3 = 0

<=> ( 5x² + 5x ) - ( 3x + 3 ) = 0

<=> 5x ( x + 1 ) - 3 ( x + 1 ) = 0

<=> ( 5x - 3 ) ( x + 1 ) = 0 

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-1\end{cases}}\)

d. ( 2x + 3 ) ( x - 3 ) - ( x - 3 ) ( x + 1 ) = ( 2 - x ) ( 3x + 1 ) + 3

<=> ( x - 3 ) ( 2x + 3 - x - 1 ) = - 3x² + 5x + 5

<=> x² - x - 6 = - 3x² + 5x + 5

<=>  - 3x² + 5x + 5 - x² + x + 6 = 0

<=> - 4x² + 6x + 11 = 0

\(\Leftrightarrow x=\frac{6\pm\sqrt{\left(-6\right)^2-4\left(4.\left(-11\right)\right)}}{2.4}\)( xài công thức bậc 2 )

\(\Leftrightarrow x=\frac{6\pm2\sqrt{53}}{8}\Leftrightarrow x=\frac{3\pm\sqrt{53}}{4}\)

Vậy \(x=\frac{3+\sqrt{53}}{4};x=\frac{3-\sqrt{53}}{4}\)