x³ + x = 0

=> x³ = 0 hoặc x = 0

* x = 0 => x = 0

* x³ = 0 => x = 0

Vậy x = 0

x = 0 nha

1; 1; 2; 4; 6; 10; 16; 26;..42..; 68.

2; 8; 14; 20; .26....; .32..... 

Học tốt!!!!!!!!!

- 42; 68
- 26; 32

giúp mình với!!

Bạn xem lại đề nhé!

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

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

\(=2a^2+2b^2+2c^2+2ab+2bc+2ac\)  (1)

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

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

\(=2a^2+2b^2+2c^2+2ab+2bc+2ca\)   (2)

Từ (1) và (2)

\(\Rightarrowđpcm\)

nhanh

Đặt \(f\left(x\right)=3x^3+ 2x^2-7x+m\)

Áp dụng định lý Bezout ta có:

\(f\left(x\right)⋮\left(3x-1\right)\Rightarrow f\left(\frac{1}{3}\right)=0\)

 \(\Leftrightarrow3\left(\frac{1}{3}\right)^3+2\left(\frac{1}{3}\right)^2-7.\frac{1}{3}+m=0\)

 \(\Leftrightarrow m-\frac{31}{18}=0\)

  \(\Leftrightarrow m=\frac{31}{18}\)

Vậy \(m=\frac{31}{18}\)để \(f\left(x\right)⋮\left(3x-1\right)\)

a) x³+y³+2x²-2xy+2y²=(x³+y³)+(2x²-2xy+2y²)=(x+y)(x²-xy+y²)+2(x²-xy+y²)=(x+y+2)(x²-xy+y²)

b) tương tự câu a 

c) (a²+b²-5)²-4(ab+2)²=(a²+b²-5)²-2²(ab+2)²

= (a²+b²-5)²-(2ab+4)²=(a²-2ab+b²-5-4)(a²+2ab+b²-5+4)

=( (a-b)² -9 )( (a+b)² -1 ) = (a-b-3)(a-b+3)(a+b-1)(a+b+1)

3x . ( x + 4 ) - x² - 4x = 0

=> 3x . ( x + 4 ) - x ( x + 4 ) = 0 

=> ( 3x - x ) ( x + 4 ) = 0

=> x ( 3 - 1 ) ( x + 4 ) = 0

=> 2x ( x + 4 ) = 0 

=> 2TH :

+)  x = 0 

+) x + 4 = 0 => x = -4

\(3x\left(x+4\right)-x^2-4x=0\)

\(3x\left(x+4\right)-x\left(x+4\right)=0\)

\(\left(3x-x\right)\left(x+4\right)=0\)

\(2x\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

Vậy \(x\in\left\{0;-4\right\}\)