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

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

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

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

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

a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)

\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)

\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)

\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)

\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)

b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)

\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)

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

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

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

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)

=>(-2x+12)(4x+12)=0

=>x=-3 hoặc x=6

b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)

=>\(x\simeq0.93\)

d: =>-4x+28+11x=-x+3x+15

=>7x+28=2x+15

=>5x=-13

=>x=-13/5

e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)

=>-9x=-3x+5

=>-6x=5

=>x=-5/6

a. x(x-5) -4x+20=0

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

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

<=>(x-5)=0 hoặc x-4=0

<=> x=5 hoặc x=4

Vậy x={4;5}

b.tương tự

c. x³-5x²+x-5 =0

<=> x²(x-5) + (x-5) = 0

<=> (x-5) (x²+1) = 0

<=> x-5=0 hoặc x²+1=0(loại vì x²=-1)

<=> x=5

vậy x=5

d. bạn kiểm tra lại đề

Tìm x :

a) \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

\(\Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-6\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vôlí\right)\\x=5\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^3+10x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x\left(x^2+10\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x^2+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x^2=-10\left(vôlí\right)|x^2\ge0\forall x\end{matrix}\right.\)