\(x^3+3x^2+\frac{2x}{x^4}+x\)

\(=x\left(x^2+3x+\frac{2}{x^4}+1\right)\)

a, \(\left(x-1\right)^2+x\left(5-x\right)=8\Leftrightarrow x^2-2x+1+5x-x^2=8\)

\(\Leftrightarrow3x+1=8\Leftrightarrow x=\frac{7}{3}\)

b, \(\left(12x^4-6x\right):6x+2x\left(2+x\right)\left(2-x\right)=7\)

Ta có : \(\left(12x^4-6x\right):6x=2x^3-1\)

\(\Leftrightarrow2x^3-1+2x\left(4-x^2\right)=7\)

\(\Leftrightarrow2x^3-1+8x-2x^3=7\Leftrightarrow x=1\)

a, \(A=x\left(x+4\right)-6\left(x-1\right)\left(x+1\right)+\left(2x-1\right)^2\)

\(=x^2+4x-6\left(x^2-1\right)+4x^2-4x+1\)

\(=x^2+4x-6x^2+6+4x^2-4x+1=7\)

b, Ko hiểu đề bài 

\(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)=\left(x+3-x+2\right)^2=5^2=25\)

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

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

\(=\left(x+3-x+2\right)^2\)   

\(=5^2\)   

\(=25\)

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Đặt \(\left(a,b,c\right)\rightarrow\left(x^3,y^3,z^3\right)\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta cần tìm GTLN của \(P=\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\)

Theo bất đẳng thức Cô-si, ta có: \(3a.a.b\le a^3+a^3+b^3=2a^3+b^3\); \(3a.b.b\le a^3+b^3+b^3=a^3+2b^3\)

Cộng theo vế hai bất đẳng thức trên, ta được: \(3ab\left(a+b\right)\le3\left(a^3+b^3\right)\Rightarrow a^3+b^3\ge ab\left(a+b\right)\)

Áp dụng, ta được: \(P=\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\le\frac{1}{xy\left(x+y\right)+1}+\frac{1}{yz\left(y+z\right)+1}+\frac{1}{zx\left(z+x\right)+1}\)\(=\frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1\)

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c = 1

Xét VP ta có:

\(\left(a-b\right)\left(a^{n-1}+a^{n-2}b+......+a^1b^{n-2}+b^{n-1}\right)\)

\(=a.a^{n-1}+a^{n-2}b.a+...+a^1b^{n-2}.a+b^{n-1}.a-a^{n-1}.b-a^{n-2}.b.b+...+ab^{n-2}.b+b^{n-1}.b\)

\(=a^n+a^{n-1}b+...+a^2b^{n-2}+b^{n-1}.a-a^{n-1}.b-a^{n-2}.b^2+...+ab^{n-2}+b^n\)

\(=a^n+b^n=VP\)( đpcm )