a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A

\(A=x^6-\left(x+1\right)^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(A=1\)

b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B

\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)

\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)

\(B=1\)

Chúc bạn học tốt!!!

Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\)    \(\forall x\)

=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\)   \(\forall x\)

=> \(20x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)  =>  \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)

                        => \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)

=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)

=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)

=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)

Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)

=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

=> \(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

=> \(\frac{1}{2}A=1-\frac{1}{20}\)

=> \(A=\frac{19}{10}\)

Thay vào ta có 

=> \(x=-\frac{19}{10}\)

mk nhầm nha bạn \(x=\frac{19}{10}\)

Ta có \(x=21\Rightarrow x-1=20\)

biểu thức B có dạng :

 \(B=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)

\(=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+3=x+3\)

Vậy \(B=21+3=24\)

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

\(=\left(2.\left(-\frac{1}{2}\right)+1\right)^2-3\left(-\frac{1}{2}-1\right)^2-\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)\)

\(=-3\left(-\frac{9}{4}\right)-\frac{1}{2}.\left(-\frac{3}{2}\right)\)

\(=\frac{27}{4}+\frac{3}{4}=\frac{31}{4}\)

còn đâu tự lm nha ! 

`F(x)=2x+10*x^3+10*(-1)+20x^6-5*x^7+5*x^5+1,5x^4-10+6x`

`F(x)=`\(2x+10x^3-10+20x^6-5x^7+5x^5+1,5x^4-10+6x\)

`F(x)= -5x^7+20x^6+5x^5+1,5x^4+10x^3+(2x+6x)+(-10-10)`

`F(x)= -5x^7+20x^6+5x^5+1,5x^4+10x^3+8x-20`

\(M=x^6-20x^5-20x^4-20x^3-20x^2-20x+3\)

\(M=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)

\(M=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x+3\)

\(M=x+3\) (1)

Thay \(x=21\)vào (1) ta được:

\(M=21+3\)

\(M=24\)

Còn câu N bạn tham khảo tại link này nha:

Câu hỏi của Hoang Linh - Toán lớp 8 | Học trực tuyến

Chúc bạn học thật tốt!

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6