Vì \(a^2+b^2+c^2=1\)

\(\Rightarrow-1\le a,b,c\le1\)

\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)

Lây cai xau trừ cai trươc được

\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)

\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)

Ta co \(VT\le0\)

Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)

\(\Rightarrow S=1\) 

=(1+4+4²) +(4³+4⁴+4⁵)+....+(4²⁰¹⁷+4²⁰¹⁸+4²⁰¹⁹⁾

=(1+4+4²)+4³(1+4+4²)+.....+4²⁰¹⁷(1+4+4²)

=(1+4+4²)(1+4³+4⁶+....+4²⁰¹⁷)

=(1+4+16)(1+4³+4⁶+.....+4²⁰¹⁷)

=21(1+4³+4⁶+...+4²⁰¹⁷) 

Vậy 21(1+4³+4⁶+.....+4²⁰¹⁷) chia hết cho 21

\(1+4+4^2+4^3+4^4+....+4^{2019}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+......+\left(4^{2017}+4^{2018}+4^{2019}\right)\)

\(=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+.....+4^{2017}\left(1+4+4^2\right)\)

\(=\left(1+4+4^2\right)\left(1+4^3+.....+4^{2017}\right)\)

\(=21\left(1+4^3+....+4^{2017}\right)\)

Mà \(21⋮21\Rightarrow21\left(1+4^3+.....+4^{2017}\right)⋮21\)

Vậy biểu thức trên chia hết cho 21(đpcm)

a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

1.The glass is broken into small pieces.

2.Then the glass is washed with a detergent liquid.

3.The mixture is melted until it becomes a liquid.

4.A long pipe is used to dip the mixture into the liquid and then the liquid is blown into intended shapes.

A long pipe is used. The mixture is dipped into the liquid. Then the liquid is blown into intended shapes.

B1:

a,\(\left(3x-2\right)\left(x-3\right)=3x^2-9x-2x+6=3x^2-11x+6\)

b,\(\left(2x+1\right)\left(x+3\right)=2x^2+6x+x+3=2x^2+7x+3\)

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

B2:

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

2)\(x\left(x+2\right)-\left(x-2\right)\left(x+4\right)=x^2+2x-\left(x^2+4x-2x-8\right)\)

\(=x^2+2x-x^2-4x+2x+8=8\)

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

=3x.x+3x.(-3)-2.x-2.(-3)

=\(3x^2\)-9x-4x+6

=\(3x^2\)+(-9x-4x)+6

=\(3x^2\)-13x+6

5) (2x+1)(x+3)=2x(x+3)+1(x+3)

=2x.x+2x.3+1.x+1.3

=\(2x^2\)+6x+1x+3

=\(2x^2\)+(6x+1x)+3

=\(2x^2\)+7x+3

6) (x-3)(3x-1)=x(3x-1)-3(3x-1)

=x.3x+x.(-1)-3.3x-3.(-1)

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

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

=\(3x^2\)-10x+3

rút gọn biểu thức

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

=\(x^2\)-x.x-x.(-1)-4.x-4.(-1)

=\(x^2\)-\(x^2\)+1x-4x+4

=(\(x^2-x^2\))+(1x-4x)+4

= -3x+4

B) x(x+2)-(x-2)(x+4)=x.x+x.2-x(x+4)+2(x+4)

=\(x^2+2x\)-x.x-x.4+2.x+2.4

=\(x^2+2x-x^2-4x+2x+8\)

=(\(x^2-x^2\))+(2x-4x+2x)+8

=8

tính giá trị biểu thức

A=3(x-2)-(2+x)(x-3)

=3.x+3.(-2)-2(x-3)-x(x-3)

=3x-6-2.x-2.(-3)-x.x-x(-3)

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

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

=4x - \(x^2\)

thay x=-8 vào biểu thức thu gọn ta được:

4.(-8)- (-8)\(^2\)

= - 32 +64

= 32

B= x(3-x)-(1+x)(1-x)

=x.3+x.(-x)-1(1-x)-x(1-x)

=3x -\(x^2\)-1.1-1 .(-x)-x.1-x.(-x)

=3x\(-x^2\)-\(1^2\)+1x-1x+\(x^2\)

=(3x+1x-1x)+(\(-x^2+x^2\))-1

=3x-1

thay x=-5 vào biểu thức thu gọn ta được:

3.(-5)-1

=-15-1

=-16

Thu gọn biểu thức

4) (3x - 2) (x - 3) 

= ( 3x² - 2x ) - ( 3x x 3 - 2 x 3 )

= 3x² - 2x - 3x x 3 + 2 x 3

= 3x² - 2x - 9x + 6

= 3x² - 11x + 6 

5) (2x + 1) (x + 3) 

= ( 2x² + 1x ) + ( 6x + 3 )

= 2x² + 1x + 6x + 3

= 2x² + 7x + 3

6) (x - 3) (3x - 1) 

= ( 3x² - 9x ) - ( x - 3 )

= 3x² - 9x - x + 3

= 3x² - 10 + 3

Rút gọn biểu thức

A) x^2 - (x + 4) (x - 1)

= x² - ( x² + 4x ) - ( x + 4 )

= x² - x² - 4x - x - 4

= -5x - 4

B) x (x + 2) - (x - 2) (x + 4)

= x² + 2x - ( x² - 2x ) + ( 4x - 8 )

= x² + 2x - x² + 2x + 4x - 8

= 8x - 8

Tính giá trị biểu thức

A = 3 (x - 2) - (2 + x) (x - 3) tại x = - 8

Thế x = -8 vào, ta có :

= 3 ( -8 -2 ) - ( 2 + -8 ) ( -8 - 3 )

= 3 x ( -10 ) - ( - 6 ) ( -11 )

= -30 - 66

= -96

B = x (3 - x) - (1 + x) ( 1 - x) tại x = - 5

Thế x = - 5 vào, ta có :

= -5 ( 3 - -5 ) - ( 1+ -5 ) ( 1 - -5 )

= -5 x 8 - (-4) x 6

= - 40 - -24

= -40 + 24

= -16

100% đúng 

hok tốt nha 

=1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1) 
=(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n) 
ta có các công thức: 
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6 
1 + 2 + 3 + ...+ n = n(n+1)/2 
thay vào ta có: 
S = n(n+1)(2n+1)/6 + n(n+1)/2 
=n(n+1)/2[(2n+1)/3 + 1] 
=n(n+1)(n+2)/3