TH1 : \(x< -2020\) 

<=> | x + 1 | + | x + 2 | + | x + 2020 | = - ( x + 1 ) - ( x + 2 ) - ( x + 2020 ) = 4x

<=> -3x - 2023 = 4x <=> -7x = 2023 <=> x = -289

TH2 : \(-2020\le x< -2\)

<=> | x + 1 |  + | x + 2 | + | x + 2020 | = - ( x + 1 ) - ( x + 2 ) + x + 2020 = 4x

<=> -x + 2017 = 4x 

<=> -5x = -2017 <=> x = 2017/5   ( = 403,4 )

TH3 : \(-2\le x< -1\)

<=> | x + 1 | + | x + 2 | + | x + 2020 | = - ( x + 1 ) + x + 2 + x + 2020 = 4x 

<=> x + 2021 = 4x <=> -3x = -2021 <=> x = 2021/3 

TH4 : \(x>-1\)

<=> | x + 1 | + | x + 2 | + | x + 2020 | = x + 1 + x + 2 + x + 2020 = 4x

<=> 3x + 2023 = 4x 

<=> -x = -2023 <=> x = 2023 

Vậy...

TH1: x ≥ 0

Khi đó \(\left|x+1\right|+\left|x+2\right|+\left|x+2020\right|=x+1+x+2+x+2020\)

                                                           \(=3x+2023=4x\)

Suy ra \(4x-3x=x=2023\) (thỏa mãn điều kiện)

TH2: x < 0

Khi đó 4x < 0 hay vế phải luôn là một số âm. Tuy nhiên vế trái luôn luôn có giá trị lớn hơn 0 nên luôn là 0 hoặc là một số dương, suy ra vô lí.

Tóm lại, x = 2023.

a) 5x(x - 1) = x - 1

    5x(x - 1) - (x - 1) = 0

    (x - 1) (5x - 1) = 0

TH1: x - 1 = 0

        x       = 1

TH2: 5x - 1 = 0

         5x      = 1

          x       = 1/5

Vay x = 1 hoac x = 1/5.

b) 2(x + 5) - x² - 5x = 0

    2(x + 5) - x(x + 5) = 0

    (x + 5) (2 - x) = 0

TH1: x + 5 = 0

        x        = -5

TH2: 2 - x = 0

              x  = 2

Vay x = -5 hoac x = 2

a, x = 1

b , x = -5 hoặc x = 2

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

__

\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

1) A=1/8; 2) B=9472; 3) 0 và 10.

1: \(A=2x^3y^4-5x\cdot x^2y^4+xy^2\cdot x^2y^2=-2x^3y^4=-2\cdot\left(-1\right)^3\cdot\dfrac{1}{16}=\dfrac{1}{8}\)

2: \(B=9x^4y^6\cdot\left(-4xy\right)+19x^3y^5\cdot\left(-2\right)x^2y^2\)

\(=-36x^5y^7-38x^5y^7\)

\(=-74x^5y^7=-74\cdot\left(-1\right)^5\cdot2^7=9472\)

3: \(f\left(-1\right)=3\cdot\left(-1\right)^4+7\cdot\left(-1\right)^3+4\cdot\left(-1\right)^2-2\cdot\left(-1\right)-2=0\)

\(f\left(1\right)=3+7+4-2-2=10\)

\(Câu8\)

\(a,A=\dfrac{1}{2}x^3\times\dfrac{8}{5}x^2=\left(\dfrac{1}{2}\times\dfrac{8}{5}\right)x^{3+2}=\dfrac{4}{5}x^5\)

b, \(P\left(0\right)=0^2-5.0+6=6\\ P\left(2\right)=2^2-5.2+6=0\)

Câu 9

\(a,A\left(x\right)+B\left(x\right)=5x^3+x^2-3x+5+5x^3+x^2+2x-3\\ =\left(5x^3+5x^3\right)+\left(x^2+x^2\right)+\left(-3x+2x\right)+\left(5-3\right)\\ =10x^3+2x^2-x+2\)

\(b,H\left(x\right)=A\left(x\right)-B\left(x\right)=5x^3+x^2-3x+5-\left(5x^3+x^2+2x-3\right)\\ =5x^3+x^2-3x+5-5x^3-x^2-2x+3\\ =\left(5x^3-5x^3\right)+\left(x^2-x^2\right) +\left(-3x-2x\right)+\left(5+3\right)\\ =-5x+8\)

\(H\left(x\right)=0\\ \Rightarrow-5x+8=0\\ \Rightarrow x=\dfrac{8}{5}\)

vậy nghiệm của đa thức là \(x=\dfrac{8}{5}\)

A(x) ở đâu

Đề có gì đó hơi sai sai 

a) Ta có: A(x)=M(x)-N(x)

\(=\dfrac{5}{2}x^2-\dfrac{1}{2}x-x^3-1-\left(-2x^3+5x^2-12+4x\right)\)

\(=\dfrac{5}{2}x^2-\dfrac{1}{2}x-x^3-1+2x^3-5x^2+12-4x\)

\(=x^3-\dfrac{5}{2}x^2-\dfrac{9}{2}x+11\)