Ta có: \(2\left(a^4+b^4\right)-\left(ab^3+a^3b+2a^2b^2\right)\)

\(=\left(a^2-b^2\right)^2+\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Ta có đpcm

Trả lời:

\(2x\left(x^2-7x-3\right)=2x^3-14x^2-6x\)

\(\left(2x^2+y^2-7xy\right).4xy^2=8x^3y^2+4xy^4-28x^2y^3\)

\(-5x^3.\left(2x^2+3x-5\right)=-10x^5-15x^4+25x^3\)

\(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-6x^5+3x^4y-3x^3y^2\)

\(\left(x^2-2x+3\right)\left(x-4\right)=x^3-4x^2-2x^2+8x+3x-12=x^3-6x^2+11x-12\)

\(\left(2x^3-3x-1\right)\left(5x+2\right)=10x^4+4x^3-15x^2-6x-5x-2=10x^4+4x^3-15x^2-11x-2\)

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

Để \(f\left(x\right)\)chia hết cho \(x+2\)thì \(f\left(-2\right)=0\)

\(\Rightarrow2.\left(-2\right)^3-3.\left(-2\right)^2+\left(-2\right)+A=0\)

\(\Leftrightarrow A=30\)

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TL

x=-1

HT 

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

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

\(=\left(x+y\right)\left(x+y+5\right)-2\left(x+y+5\right)\)

\(=\left(x+y-2\right)\left(x+y+5\right)\)

\(\left(2m-a\right)^2+\left(2m-b\right)^2+\left(2m-c\right)^2+\left(2m-d\right)^2+\left(2m-e\right)^2\)

\(=4m^2-4ma+a^2+4m^2-4mb+b^2+4m^2-4mc+c^2+4m^2-4md+d^2+4m^2-4me+e^2\)

\(=20m^2-4m\left(a+b+c+d+e\right)+a^2+b^2+c^2+d^2+e^2\)

\(=20m^2-4m.5m+a^2+b^2+c^2+d^2+e^2\)

\(=a^2+b^2+c^2+d^2+e^2\)

a) 

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

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

\(=4x+1\)

b) 

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

\(=x^3-6x^2+12x-8-x^2+8\)

\(=-6x^2+12x\)