a, \(16x^2-9\left(x+1\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)

c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)

1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

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

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

2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3: Ta có: \(5x^3-20x=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)

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

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

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

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

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

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

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

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

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

a) x(x - 2) + x -2 =0

<=>x.(x-2)+(x-2)=0

<=>(x-2)(x+1)=0

<=>x-2=0 hoặc x+1=0

<=>x=2 hoặc x=-1

b) 5x(x - 3) - x +3 =0

<=>5x(x-3)-(x-3)=0

<=>(x-3)(5x-1)=0

<=>x-3=0 hoặc 5x-1=0

<=>x=3 hoặc x=1/5

c) x + 4 - x(x+4)=0

 <=>(x+4)-x.(x+4)=0

<=>(x+4)(1-x)=0

<=>x+4=0 hoặc 1-x=0

<=>x=-4 hoặc x=1

Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)

\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)

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

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)

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

\(\Leftrightarrow2x+1=0\)

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

\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

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

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

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

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

\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)

\(=25x^2+45x+15+8+10x-40x-50x^2\)

\(=-25x^2+15x+23\)

\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)

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

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

\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)

\(=x^3+y^3\)

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

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

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

....

1) x = 0

2) x = 2

3) không biết (thông cảm)

4) x > 0

5) x < 0

Đặt Q là thương của phép chia . Vì đây là phép chia hết nên ta có phương trình

5x⁴+5x³+x²+11x+a = (x²+x+b)Q . Mà vế trái là đa thức bậc 4 nên khi chia cho đa thức bậc 2 thì thương có dạng Q = mx²+nx+h 

( với m,n,h là hệ số của đa thức )

=>  5x⁴+5x³+x²+11x+a = (x²+x+b)(mx²+nx+h)

<=>5x⁴+5x³+x²+11x+a = mx⁴+ nx³ + hx² + mx³ + nx² + hx + bmx² + bnx + bh

                                   = mx⁴ + (m+n)x³ + (h+n+bm)x² + (h+bn)x + bh

Mà theo nguyên tắc hai vế bằng nhau thì hệ số của bậc nào bằng hệ số bậc cùng bậc bên vế kia .

=> m = 5 

     m+n = 5 => n = 0

     h+bn = 11 => h = 11

     h+n+bm = 1 => b = -2

     bh = a = -22

Vậy a = -22 ; b = -2 ; Q = 5x²+11

                                                         x⁴-30x²+31x-30 = 0 

<=> x⁴ + ( x³ - x³ ) + ( x² - x² - 30x² ) + ( 30x + x ) -30 = 0

<=> ( x⁴ + x³ - 30x² ) + ( -x³ - x² + 30x ) + ( x² + x - 30 ) =0

<=> x².( x² + x - 30 ) - x.( x² + x - 30 ) + ( x² + x - 30 )  = 0

<=>                       ( x² + x - 30 )( x² - x + 1 )               = 0 

<=>                       ( x² + x - 30 )( x - 5 )( x + 6 )           = 0 

Vì  x² + x - 30 =  x² + x + \(\frac{1}{4}\) - \(\frac{121}{4}\) = ( x + \(\frac{1}{2}\) )² - \(\frac{121}{4}\) \(\ge\)- \(\frac{121}{4}\) 

=> x - 5 = 0 hoặc x + 6 = 0 

=>      x = 5 hoặc      x = -6

Vậy tập nghiệm S = { -6 ; 5 }

b) \(x^4-30x^2+31-30=0\)

\(\Leftrightarrow x^4+x-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-5=0\\x^2-x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=5\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\text{loai}\right)\end{cases}}}\)

\(\Rightarrow x\in\left\{-6;5\right\}\)

a,x(x-2)+x-2=0

⇔ (x-2)(x+1)=0

⇔ x=2;x=-1

b,x³+x²+x+1=0

⇔ x²(x+1)+x+1=0

⇔ (x+1)(x²+1)=0

⇔ x=-1