Ta có: \(2020=x\Rightarrow2019=x-1\)

Thay vào ta được:

\(D=x^{2020}+\left(x-1\right)^{2019}+\left(x-1\right)^{2018}+...+\left(x-1\right)x+1\)

\(D=x^{2020}+x^{2020}-x^{2019}+x^{2019}-x^{2018}+...+x^2-x+1\)

\(D=2x^{2020}-x+1\)

\(D=2\cdot2020^{2020}-2020+1\)

Bạn xem lại đề nhé

x = 2020 => 2019 = x - 1

Thế vào D ta được

D = x²⁰²⁰ + ( x - 1 )x²⁰¹⁹ + ( x - 1 )x²⁰¹⁸ + ... + ( x - 1 )x + 1

= x²⁰²⁰ + x²⁰²⁰ - x²⁰¹⁹ + x²⁰¹⁹ - x²⁰¹⁸ + ... + x² - x + 1

= 2x²⁰²⁰ - x + 1 

= 2.2020²⁰²⁰ - 2020 + 1 

= 2.2020²⁰²⁰ - 2019 ( chắc đề sai (: )

\(\left(x-y\right)^2+\left(x-y\right)^2\)

\(=2\left(x-y\right)^2\)

Khai triển?

\(\left(x-y\right)^2+\left(x-y\right)^2\)

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

\(=2x^2-4xy+2y^2\)

a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)

b) \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

a) x² + x + 1 = ( x² + x + 1/4 ) + 3/4 = ( x + 1/2 )² + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )

b) 4x² - 2x + 1 = 4( x² - 1/2x + 1/16 ) + 3/4 = 4( x - 1/4 )² + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )

c) x⁴ - 3x² + 9 (*)

Đặt t = x²

(*) <=> t² - 3t + 9 = ( t² - 3t + 9/4 ) + 27/4 = ( t - 3/2 )² + 27/4 = ( x² - 3/2 )² + 27/4 ≥ 27/4 > 0 ∀ x ( đpcm )

d) x² + y² - 2x - 4y + 6 = ( x² - 2x + 1 ) + ( y² - 4y + 4 ) + 1 = ( x - 1 )² + ( y - 2 )² + 1 ≥ 1 > 0 ∀ x, y ( đpcm )

e) x² + y² - 2x - 2y + 2xy + 2 = ( x² + 2xy + y² - 2x - 2y + 1 ) + 1

                                              = [ ( x² + 2xy + y² ) - ( 2x + 2y ) + 1 ] + 1 

                                              = [ ( x + y )² - 2( x + y ) + 1² ] + 1

                                              = ( x + y - 1 )² + 1 ≥ 1 > 0 ∀ x, y ( đpcm )

a) \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)

b) \(4x^2-2x+1=4\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{3}{4}=4\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)

c) \(x^4-3x^2+9=\left(x^4-3x^2+\frac{9}{4}\right)+\frac{27}{4}=\left(x^2-\frac{3}{2}\right)^2+\frac{27}{4}>0\left(\forall x\right)\)

d) \(x^2+y^2-2x-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\left(\forall x,y\right)\)

e) \(x^2+y^2-2x-2y+2xy+2\)

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

\(=\left(x+y-1\right)^2+1>0\left(\forall x,y\right)\)

a) x² + 2x - 3 = x² - x + 3x - 3 = x( x - 1 ) + 3( x - 1 ) = ( x - 1 )( x + 3 )

b) x² - 2x - 15 = x² + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )

c) x² - 2x - 48 = x² + 6x - 8x - 48 = x( x + 6 ) - 8( x + 6 ) = ( x + 6 )( x - 8 )

d) 4x² + 4x - 15 = ( 4x² + 4x + 1 ) - 16 = ( 2x + 1 )² - 4² = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )

e) 3x² - 7x + 2 = 3x² - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )

f) 4x² - 5x + 1 = 4x² - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )

b) x² - 10x + 16 = x² - 2x - 8x + 16 = x( x - 2 ) - 8( x - 2 ) = ( x - 2 )( x - 8 )

c) x² + 6x + 8 = x² + 2x + 4x + 8 = x( x + 2 ) + 4( x + 2 ) = ( x + 2 )( x + 4 )

d) x² - 8x + 15 = x² - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )

e) x² - 8x - 9 = x² + x - 9x - 9 = x( x + 1 ) - 9( x + 1 ) = ( x + 1 )( x - 9 )

f) x² + 14x + 48 = x² + 6x + 8x + 48 = x( x + 6 ) + 8( x + 6 ) = ( x + 6 )( x + 8 )

\(2\left(x^2+y^2\right)=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)\(=\left(x+y\right)^2+\left(x-y\right)^2\)

Bài làm :

\(2.\left(x^2+y^2\right)\)

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

\(=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)

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

-> đpcm

Học tốt

2x² - x - 6 = 0

<=> 2x² - 4x + 3x - 6 = 0

<=> 2x ( x - 2 ) + 3 ( x - 2 ) = 0

<=> ( 2x + 3 ) ( x - 2 ) = 0

<=> \(\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

2x² - x - 6 = 0

<=> 2x² + 3x - 4x - 6 = 0

<=> 2x( x + 3/2 ) - 4( x + 3/2 ) = 0

<=> ( x + 3/2 )( 2x - 4 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{2}=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

a( a - b ) + b( b - c ) + c( c - a ) = 0

<=> a² - ab + b² - bc + c² - ca = 0

Nhân 2 vào từng vế 

<=> 2( a² - ab + b² - bc + c² - ca ) = 2.0

<=> 2a² - 2ab + 2b² - 2bc + 2c² - 2ca = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) = 0

<=> ( a - b )² + ( b - c )² + ( c - a )² = 0 (*)

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

=> đpcm

a ( a - b ) + b ( b - c ) + c ( c - a ) = 0

<=> a² + b² + c² - ab - bc - ca = 0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) = 0

<=> ( a - b )² + ( b - c )² + ( c - a )² = 0

Mà ( a - b )² + ( b - c )² + ( c - a )² \(\ge\)0\(\forall\)a ; b ; c

Dấu "=" xảy ra <=> a = b = c ( đpcm )