• Ta có : \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{4}{2}=-2\)

• Ta có ; \(\left(a^2+b^2+c^2\right)^2=16\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=16\)

\(\Leftrightarrow a^4+b^4+c^4=16-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

Mặt khác : \(\left(ab+bc+ac\right)^2=4\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=4\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=4\)

\(\Rightarrow a^4+b^4+c^4=16-2.4=8\)

Ta có a + b + c = 0 

<=> (a + b + c)² = 0

<=> a² + b² + c² + 2(ab + bc + ca) = 0 

<=> ab + bc + ca = \(-\frac{1}{2}\)

=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Lại có a² + b² + c² = 1

=> (a² + b² + c²)² = 1

<= > a⁴ + b⁴ + c⁴ + 2[(ab)² + (bc)² + (ca)²] = 1 

<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)

<=> \(a^4+b^4+c^4=\frac{1}{2}\)

Từ a + b + c = 0 => ( a + b + c )² = 0 <=> a² + b² + c² + 2ab + 2bc + 2ca = 0

<=> ab + bc + ca = -1/2 => ( ab + bc + ca )² = 1/4

<=> a²b² + b²c² + c²a² + 2ab²c + 2bc²a + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc( a + b + c ) = 1/4

<=> a²b² + b²c² + c²a² = 1/4 ( vì a + b + c = 0 )

Từ a² + b² + c² = 1 => ( a² + b² + c² )² = 1 <=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 1

<=> a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 1 

<=> a⁴ + b⁴ + c⁴ + 1/2 = 1 <=> a⁴ + b⁴ + c⁴ = 1/2

Vậy A = 1/2

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

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

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)

\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)

b: Để A>0 thì x-3>0

hay x>3

bằng 1/2 bạn ơi