\(\left(a+b+c\right)=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Rightarrow2ab+2bc+2ac=-2\)

\(\Rightarrow ab+bc+ac=-1\Rightarrow\left(ab+bc+ac\right)^2=1\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2+2abc\left(a+b+c\right)=4\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+0=4\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=4\)

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

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

Bn làm phần kết quả nhé

\(C=x^2-5y^2+x-3y\)

\(C=x^2-y^2-4y^2+x-3y\)

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

\(C=x-y-4y^2+x-3y\)

\(C=-4y^2+2x-4y\)

\(C=-4y^2-4y-1+2x+1\)

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

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

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

bn làm tiếp nha

giá trị âm nhá

A = 2x - x² - 2 

= -(x² - 2x + 2)

= -(x² - 2x +  1 + 1)

= -(x² - 2x + 1) - 1

= -(x - 1)² - 1 

Vì (x - 1)² \(\ge0\forall x\)

=> -(x - 1)² \(\le0\forall x\)

Vậy A = -(x - 1)² - 1 \(\le1< 0\forall x\)

\(a=2x-x^2-2\)

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

\(a=-x^2+2x-1-1\)

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

Dấu "=" xảy ra khi x = 1

Vậy x luôn âm

1) \(x^2-2xy+y^2-xz+yz\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)

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

\(\Leftrightarrow\left(x-y\right)\left(x-y-z\right)\)

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

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)\)

\(a,x^2-2xy+y^2-xz+yz\)

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

\(=\left(x-y\right)\left(x-y-z\right)\)

\(b,x^2-y^2-x+y\)

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

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

1) (a + b)² - (a - b)² = 4ab

VT = (a + b) ² - ( a - b ) ² = ( a² + 2ab + b²) - (a² - 2ab + b² )  = a² + 2ab + b² - a² + 2ab - b² = 4ab = VP (đpcm)

2) (a + b) ² + (a - b)² = 2(a² + b² )

VT = (a + b)² + (a - b)² = a² + 2ab + b² + a² - 2ab + b² = 2a² + 2b² = 2 (a² + b²) = VP (đpcm)

3) (a + b)² - 4ab = (a - b)²

VT = (a + b)² - 4ab = a² + 2ab + b² - 4ab = a² - 2ab + b² = (a - b)² = VP (đpcm)

4) (a - b)² + 4ab = (a + b)²

VT = (a - b)² + 4ab = a² - 2ab + b² + 4ab = a² + 2ab + b² = (a + b)² = VP (đpcm)

5) a³ + b³ = (a + b)³ - 3ab (a + b)

VP = (a + b)³ - 3ab (a + b) = a³ + 3a²b + 3ab² + b³ - 3a²b - 3ab² = a³+ b³ = VT (đpcm)

6) a³ - b³ = (a - b)³ + 3ab (a - b)

VP = (a - b)³ + 3ab (a - b) = a³ - 3a²b + 3ab² - b³ + 3a²b - 3ab² = a³- b³ = VT (đpcm)

7) a³ + b³ + c³ - 3abc = ( a + b + c) ( a² + b² + c² - ab - bc - ac )

VP =  (a + b + c) (a² + b² + c² - ab - bc - ac)

     = a³ + ab²  + ac² - a²b - abc - a²c + a²b + b³ + bc² - ab² - b²c - abc + a²c + b²c + c³ - abc - bc² - ac² 

     = a³ + b³ + c³ - 3abc = VT (đpcm) 

câu 7 mk sửa đề lại xíu nhea !!!

có j sai xót mong m.n bỏ qa cho ☺♥

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