a) Từ \(a+b+c=0\Rightarrow a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=0\)

\(\Rightarrow ab+bc+ca=-1 \) (do \( a^{2}+b^{2}+c^{2}=2\))

\(\Rightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2(a^{2}bc+ab^{2}c+abc^{2})=1\)

\(\Rightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)=1\)

\(\Rightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=1\) (1)

(do a+b+c=0)

Ta có: \(a^{2}+b^{2}+c^{2}=2\)

\(\Rightarrow a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})=4\) (20

\((1)+(2)\Rightarrow a^{4}+b^{4}+c^{4}=2\)

b) Tương tự.

tới bước 3 biến đỗi sao thế

\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)

\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)

\(A=-12x\)

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

\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)

\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)

\(B=-3x-12\)

Câu C tương tự.

Chúc bạn học tốt!!!

A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)

A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)

A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)

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

A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)

A = \(-4x^2-12x-8+4x^2+8=-12x\)

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

B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)

B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)

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

a: \(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\ge2\)

b: \(6x-x^2+1\)

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

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10\le10\)

Giải:

Ta có:

\(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)

\(=\left(5a-3b\right)^2-\left(8c\right)^2\)

Mà \(a^2-b^2=4c^2\) nên:

\(VT=25^2-30ab+9b^2-16\left(a^2-b^2\right)\)

\(=9a^2-30ab+25b^2\)

\(=\left(3a-5b\right)^2=VP\) (Đpcm)

Ta có:
A = (5a – 3b + 8c)(5a – 3b –8c)
= (5a –3b)² – (8c)²
= (25a² – 30ab +9b²) – 64c²
Mà theo đề thì 4c² = a² –b²
Nên ta suy ra:
A = (25a² – 30ab +9b²) – 16(a² –b²)
= 9a² –30ab +25b²
= (3a –5b)²

a, Theo bài ra ta có:

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

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

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

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

b, theo bài ra ta có:

\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)

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

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

c,Theo bài ra ta có:

\(=x^3+5x^2+3x^2+15x+2x+10\)

\(=x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2+3x+2\right)\)

\(=\left(x+5\right)\left(x^2+x+2x+2\right)=\left(x+5\right)\left(x\left(x+1\right)+2\left(x+1\right)\right)\)

\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)

CHÚC BẠN HỌC TỐT...........

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

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

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

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

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

= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

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

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

b) \(x^3-5x^2+3x+9\)

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

= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2-6x+9\right)\)

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

c) \(x^3+8x^2+17x+10\)

= \(x^3+x^2+7x^2+7x+10x+10\)

= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+7x+10\right)\)

= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)

= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)

= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

d) \(x^3-3x^2+6x+4\)

Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:

\(x^3+3x^2+6x+4\)

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

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

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

Học tốt nhé :))

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

\(=2a^3-6abc+2b^3+2c^3\)

Khỏi ghi lại đề:

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

\(=2a^3+2b^3+2c^3-6abc\)

Bất đẳng thức à

ủa nhưng mà thỏa mãn cái gì mới c.m mấy cái kia chứ

giải bài 1 dùm mk đi

Bài 1 sai đề nhé.

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

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

= \(z^2\)

Ta có:(x + y + z)² - 2(x + y + z) (x + y) + (x + y)²

=[(x+y+z)-(x+y)]²=z²