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2. Đặt c + d = x
Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)
\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)
Câu 4:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)
\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)
\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )
\(\Rightarrow a-b=0,b-c=0,a-c=0\)
Thay vào A ta tính được A = 0
\(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Rightarrow\left(2x^2+4xy+2y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
Khi đó: \(A=\left(-1+1\right)^{2014}+\left(-1+2\right)^{2015}+\left(1-1\right)^{2016}\)
\(=0+1+0=1\)
a) \(x^7+x^5+1\)
\(=x^7-x+x^5-x^2+x^2+x+1\)
\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)]
\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+x^2\left(x-1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[x\left(x^4-x^3+x-1\right)+x^3-x^2+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
b) \(x^5-x^4-1\)
\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)