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

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

\(\Leftrightarrow x\left(x-2\right)=0\) (do \(x^2+10>0;\forall x\))

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

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

`x^4-2x^3+10x^2-20x=0`

`<=>x^3(x-2)+10x(x-2)=0`

`<=>(x^3+10x)(x-2)=0`

`<=>x(x^2+10)(x-2)=0`

`<=>`$\left[\begin{matrix} x=0\\ x^2+10=0\\x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\ x^2=-10 \ \rm(loại) \\x=2\end{matrix}\right.$

Vậy `S={0;2}`

Ta có: \(x^4-2x^3+10x^2-20x=0\)

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

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

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

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

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

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

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

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

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-5)^2-49=0

<=>(x-5)²-7²

<=>(x-5-7)(x-5+7)=0

<=>(x-12)(x+2)=0

<=>x-12=0 hoặc x+2=0

<=>x=12 hoặc x=-2

vậy x=12 hoặc x=-2

b/ (x+11)^2=121

<=>(x+11)²-121=0

<=>(x+11)²-11²=0

<=>(x+11-11)(x+11+11)=0

<=>x(x+22)=0

<=>x=0 hoặc x+22=0

<=>x=0 hoặc x=-22

vậy x=0 hoặc x=-22

c/ x.(x+7)-6x-42=0

<=>x²+7x-6x-42=0

<=>x²+x-42=0

<=>x²-6x+7x-42=0

<=>x(x-6)+7(x-6)=0

<=>(x-6)(x-7)=0

<=>x-6=0 hoặc x-7=0

<=>x=6 hoặc x=7

vậy x=6;7

d/ x^4-2x^3+10x^2-20x=0

<=>x³(x-2)+10x(x-2)=0

<=>(x-2)(x³+10x)=0

<=>(x-2)x(x²+10)=0

<=>x-2=0 hoặc x=0 hoặc x²+10=0

<=>x=2 hoặc x=0 hoặc x²=-10(vô lí)

vậy x=2;0

a)(x-5)²-49=0

<=>(x-5-7)(x-5+7)=0

<=>(x-12)(x+2)=0

<=>x-12=0 hoặc x+2=0

<=>x=12 hoặc x=-2

b)(x+11)²=121

<=>(x+11)²-121=0

<=>(x+11-11)(x+11+11)=0

<=>x(x+22)=0

<=>x=0 hoặc x+22=0

<=>x=0 hoặc x=-22

c)x(x+7)-6x-42=0

<=>x(x+7)-(6x+42)=0

<=>x(x+7)-6(x+7)=0

<=>(x+7)(x-6)=0

<=>x+7=0 hoặc x-6=0

<=>x=-7 hoặc x=6

d)x⁴-2x³+10x²-20x=0

<=>x(x³-2x²+10x-20)=0

<=>x[(x³-2x²)+(10x-20)]=0

<=>x[x²(x-2)+10(x-2)]=0

<=>x(x-2)(x²+10)=0

Do x²>0=>x²+10>0

=>x(x-2)=0

<=>x=0 hoặc x-2=0

<=>x=0 hoặc x=2 

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

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

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

\(\Rightarrow\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

* \(x^3-16x=0\)

\(\Rightarrow x\left(x^2-16\right)=0\)

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

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

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

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

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

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

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

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

\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)

\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)

\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)

Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:

\(E=5x.0+105=105\)

1)

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

\(\Rightarrow\left(x^2+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(sai\right)\\x=5\end{cases}}\)\(KL:x=5\)

b) \(x^4-2x^3+10x^2-20x=0\Rightarrow x^3.\left(x-2\right)+10x\left(x-2\right)\)

\(\Rightarrow\left(x-2\right).\left(x^3+10x\right)\Rightarrow\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x\left(x^2+10\right)=0\Rightarrow x=0\end{cases}}\)

Vì nếu x² + 10 = 0 => x² = -10 ( sai )

Vậy...

1, x(x - 5) - 4x + 20 = 0

=> x(x - 5) - 4(x - 5) = 0

=> (x - 4)(x - 5) = 0

=> x - 4 = 0 hoặc x - 5 = 0

=> x = 4 hoặc x = 5

=> x thuộc {4; 5}

2, 3(x + 1) + x(x + 1) 

= (3 + x)(x + 1)

3, 2x³ + x = 0

=> x(2x² + 1) = 0

=> x = 0 hoặc 2x² + 1 = 0

=> x = 0 hoặc 2x² = -1

=> x = 0 hoặc x² = -1/2 (vô lí vì x² > hoặc = 0 với mọi x)

=> x = 0

4, x³ - 16x = 0

=> x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

=> x = 0 hoặc x² = 16

=> x = 0 hoặc x = 4 hoặc x = -4

=> x thuộc {-4; 0; 4}

5, x² + 6x = -9

=> x² + 6x + 9 = 0

=> x² + 2.3.x + 3² = 0

=> (x + 3)² = 0

=> x + 3 = 0

=> x = -3

6, x⁴ - 2x³ + 10x² - 20x = 0

=> x²(x² + 10) - 2x(x² + 10) = 0

=> (x² + 2x)(x² + 10) = 0

=> x(x +2)(x² + 10) = 0

-TH1: x = 0

-TH2: x + 2 = 0 => x = -2

-TH3: x² + 10 = 0 => x² = -10 (vô lí vì x² > hoặc = 0 với mọi x)

=> x thuộc {0; -2}

7, (2x - 3)² = (x + 5)²

-TH1: 2x - 3 = x + 5

=> x = 8

- TH2: - 2x + 3 = x + 5

=> -3x = 2

=> x = \(\frac{-2}{3}\)

- TH3: 2x - 3 = - x - 5

=> 3x = -2

=> x = \(\frac{-2}{3}\)

- TH4: - 2x + 3 = - x - 5

=> -x = -8

=> x = 8`

=> x thuộc {\(\frac{-2}{3}\); 8}