1) Ta có: \(2\left(x-y\right)+\left(x-y\right)^2+\left(y-x\right)^2\)

\(=2\left(-3-1000\right)+\left(-3-1000\right)^2+\left(3+1000\right)^2\)

\(=-2006+1006009+1006009\)

\(=2010012\)

2) \(x^3+12x^2+48x+64\)

\(=x^3+3.x^2.4+3.x.4^2+4^3\)

\(=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

3) \(x^3-6x^2+12x-8\)

\(=x^3-3.x^2.2+3.x.2^2-2^3\)

\(=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

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

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

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

= \(2\left(x-y\right)\)

Thay x = -3,y = 1000 vào ta có : 2(x - y) = 2(-3 - 1000) = 2.(-1003) = -2006

\(x^3+12x^2+48x+64\)

\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3=\left(x+4\right)^3\)

Thay x = 6 vào ta có : (6 + 4)³ = 10³ = 10000

\(x^3-6x^2+12x-8=x^3-3x^2\cdot2+3x\cdot2^2-2^3\)

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

Thay x = 22 vào ta có : (22 - 2)³ = 20³ = 8000

Khoảng cách từ TĐ tới MT là 

3.10^8.1,28=384000000m=384000km

a) \(A=x^2-4x-2=\left(x^2-4x+4\right)-6=\left(x-2\right)^2-6\ge-6\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy Min(A) = -6 khi x = 2

b) \(B=\left(x-1\right)\left(2x+3\right)-12\)

\(B=2x^2+x-3-12\)

\(B=2\left(x^2+\frac{x}{2}+\frac{1}{16}\right)-\frac{121}{8}\)

\(B=2\left(x+\frac{1}{4}\right)^2-\frac{121}{8}\ge-\frac{121}{8}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(2\left(x+\frac{1}{4}\right)^2=0\Rightarrow x=-\frac{1}{4}\)

Vậy \(Min_B=-\frac{121}{8}\Leftrightarrow x=-\frac{1}{4}\)

A = x² - 4x - 2

= ( x² - 4x + 4 ) - 6

= ( x - 2 )² - 6 ≥ -6 ∀ x

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinA = -6 <=> x = 2

B = ( x - 1 )( 2x + 3 ) - 12

= 2x² + x - 3 - 12

= 2x² + x - 15

= 2( x² + 1/2x + 1/16 ) - 121/8

= 2( x + 1/4 )² - 121/8 ≥ -121/8 ∀ x

Đẳng thức xảy ra <=> x + 1/4 = 0 => x = -1/4

=> MinB = -121/8 <=> x = -1/4

 Khoảng cách từ mặt trời đến trái đất là 149.600.000 km. Tốc độ ánh sáng là gần 300.000 km/giây. Do đó, ánh sáng Mặt Trời mất khoảng 499 giây (8 phút 19 giây) để đến Trái Đất.

\(\text{Tốc độ ánh sáng}\approx300000km\text{/}s\)

\(\text{Thời gian ánh sáng đi từ Mặt Trời đến Trái Đất là: }\)

\(149600000:300000\approx500\left(\text{giây}\right)\)

2( x + 1 )( x - 2 ) - ( x - 3 )( x + 4 ) = 7

<=> 2( x² - x - 2 ) - ( x² + x - 12 ) = 7

<=> 2x² - 2x - 4 - x² - x + 12 = 7

<=> x² - 3x + 8 = 7

<=> x² - 3x + 8 - 7 = 0

<=> x² - 3x + 1 = 0

<=> ( x² - 3x + 9/4 ) - 5/4 = 0

<=> \(\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=0\)

<=> \(\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=0\)

<=> \(\left(x-\frac{3+\sqrt{5}}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)=0\)

<=> \(\orbr{\begin{cases}x-\frac{3+\sqrt{5}}{2}=0\\x-\frac{3-\sqrt{5}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{cases}}\)

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

=> \(2\left[x\left(x-2\right)+1\left(x-2\right)\right]-x\left(x+4\right)+3\left(x+4\right)=7\)

=> \(2\left(x^2-2x+x-2\right)-x^2-4x+3x+12=7\)

=> \(2x^2-4x+2x-4-x^2-4x+3x+12=7\)

=> \(\left(2x^2-x^2\right)+\left(-4x+2x-4x+3x\right)+\left(-4+12\right)=7\)

=> \(x^2-3x+8=7\)

=> \(x^2-3x=-1\)

=> \(x^2-3x+1=0\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}-\frac{\sqrt{5}}{2}\\x=\frac{\sqrt{5}}{2}+\frac{3}{2}\end{cases}}\)

=7.8+7.9

=7. (8+9)

=7.17

=119

k mình nha

C = ( x² - 2x + 4 )( x + 2 ) - ( x - 2 )³ - 6( x - 1 )( x + 1 )

= x³ + 8 - ( x³ - 6x² + 12x - 8 ) - 6( x² - 1 )

= x³ + 8 - x³ + 6x² - 12x + 8 - 6x² + 6

= -12x + 22

Với x = 11/6

C = -12.11/6 + 22

    = -22 + 22 

    = 0

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

\(C=x^3+2^3-\left(x-2\right)^3-6\left(x^2-1\right)\)

\(C=x^3+2^3-\left(x^3-6x^2+12x-8\right)-6x^2+6\)

\(C=x^3+8-x^3+6x^2-12x+8-6x^2+6\)

\(C=22-12x\)

Thay x = 11/6 vào ta có : \(22-12\cdot\frac{11}{6}=22-22=0\)

 .\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

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

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

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

=\(\left(a+c-b\right)\left[a\left(c-b+a\right)+b\left(a+b+c\right)+c\left(a-b-c\right)\right]\)

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

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

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

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

X^3-3x^2+3x^2+3x-1

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

=x^3+3x-1

\(2.\left(3x-1\right).\left(2x-5\right)-\left(4x-1\right).\left(3x-2\right)\)

\(\Leftrightarrow6x-2.\left(2x-5\right)-12x^2-8x-3x+2\)

\(\Leftrightarrow12x^2-30x-4x+10-12x^2-11x+2\)

\(\Leftrightarrow-45x+12\)

2( 3x - 1 )( 2x - 5 ) - ( 4x - 1 )( 3x - 2 )

= 2( 6x² - 17x + 5 ) - ( 12x² - 11x + 2 )

= 12x² - 34x + 10 - 12x² + 11x - 2

= -23x + 8