\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

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

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

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

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

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

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

(x-4)(2x+6)=0

=>\(\left[{}\begin{matrix}x-4=0\\2x+6=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

Ta có :  (x - 4)(2x + 6) = 0

=> Một trong hai số phải bằng 0 

Nếu x - 4 = 0 ⇒ x = 4

Nếu 2x + 6 = 0 ⇒ x = -3

Vậy x = 4 hoặc -3

a ) 219 - 7 . ( x + 1 ) = 100

              7 . ( x + 1 ) = 119

                     x + 1   = 17

                           x   = 18

b ) ( 3x - 6 ) . 3 = 3 ⁴

       3x - 6        = 27

            3x        = 33

              x        = 11

a: \(\Leftrightarrow8x^2+16x+14x+7=0\)

=>(2x+1)(8x+7)=0

=>x=-1/2 hoặc x=-7/8

b: \(=x^3-x-6x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

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

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

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

a. 3/5 + x = 5/7

x = 5/7 - 3/5

x = 4/35

b. x x 4/11= 1/3

x = 1/3 : 4/11

x = 11/12

3/5+x=5/7
       x= 5/7 - 3/5
       x=     4/35

X x 4/11= 1/3
         X  = 1/3 : 4/11
         X  =      11/12

a)  x 2  + 2x + 1.            b) x + 3.         c)  x 2  – x + 1.

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

\(\left(3-6x\right):\left(x-3\right)-\left(3-6x\right):\left(x-3\right)=0\)

\(0=0\)

vậy \(x\in\varnothing\)

thank you !

Theo mình thì x\(\ne\)3 thì hai vế bằng nhau

a) Thu gọn và sắp xếp:
\(P\left(x\right)=2x^3-9x^2+5-4x^3+7x\)

\(P\left(x\right)=\left(2x^3-4x^3\right)-\left(9x^2+2x^2\right)+7x+5\)

\(P\left(x\right)=-2x^3-11x^2+7x+5\)

b) Thay x=1 vào đa thức P(x) ta được:

\(P\left(x\right)=\left(-1\right)^4-\left(-1\right)^3-\left(-1\right)-2=1\)

Ta có: \(x^3-7x-6=0\)

\(\Leftrightarrow x^3-x-6x-6=0\)

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

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

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

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

\(x^3-7x-6=0\)

\(\Leftrightarrow x^3-x-6x-6=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left[x\left(x-1\right)-6\right]=0\)

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

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

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

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

Vậy...