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...

b: \(\left(x^2+4\right)^2-16x^2\)

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

\(=\left(x-2\right)^2\cdot\left(x+2\right)^2\)

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

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

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

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

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

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

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

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

x3+3x2-10x=0

=>x(3+3.2-10)=0

=>x=0

x3-5x2-14x=0

=>x(3-5.2-14)=0

=>x=0

x3+5x2-24x=0

=>x(3+5.2-24)=0

=>x=0

Câu a)

\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)

\(\Rightarrow x=0;x=5;x=2\)

Câu b:

\(x^3-5x^2-14x=0\Rightarrow x\left(x^2-5x-14\right)=0\Rightarrow x\left(x^2+2x-7x-14\right)=0\Rightarrow x\left(x\left(x+2\right)-7\left(x+2\right)\right)=0\Rightarrow x\left(x-7\right)\left(x+2\right)=0\)

\(\Rightarrow x=0;x=7;x=-2\)

19.(2+3+4−5+6−7)2−9.(7x−2)=0

19.32−9(7x−2)=0

19.9−9(7x−2)=0

171−9(7x−2)=0

9(7x-2)=171

7x−2=171:9

7x−2=19

7x=19+2

7x=21

x=21:7

x=3x=3

Vậy x=3

Ta có |7x + 1| - |5x + 6| = 0

<=> |7x + 1| = |5x + 6| 

<=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=5\\12x=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)

Vậy  \(x\in\left\{\frac{5}{2};-\frac{7}{12}\right\}\)

a)3x^3-8x^2-2x+4

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

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

=(x^2-2x-2)(3x-2).đến đây cậu tự làm nha

b)x^3-4x^2+7x-6

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

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

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

.đến đây cậu tự làm nha

c)2x^3-9x+2

=2x^3-4x^2+4x^2-8x-x+2

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

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

.đến đây cậu tự làm nha

\(\left|7x+1\right|-\left|5x+6\right|=0\) <=> \(\left|7x+1\right|=\left|5x+6\right|\)

<=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\) <=> \(\orbr{\begin{cases}2x=5\\12x=-7\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)

a, x^3 - x^2 - x + 1 = 0

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

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

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

Vậy x= +- 1

b, 3x^2 - 3xy + 5y - 5x = 0

3x (x-y) - 5(x-y) =0

(3x -5)(x-y) =0

(làm tương tự như bài trên)

3^2 là thế nào cơ bạn 

\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)

\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

x=5 và 0