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 

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

= [ ( 3x² + 1 ) - 2x ][ ( 3x² + 1 ) + 2x ] - ( 3x² + 1 )²

= ( 3x² + 1 )² - 4x² - ( 3x² + 1 )²

= -4x²