+) Với \(n=1\Rightarrow B=3\) là SNT

+) Với \(n>1\Rightarrow B>3\)

Ta có: \(B=\left(n^{1988}-n^2\right)+\left(n^{1987}-n\right)+\left(n^2+n+1\right)\)

Có \(n^{1986}-1=\left[\left(n^3\right)^{662}-1\right]⋮n^3-1\)

Mà \(n^3-1=\left(n-1\right)\left(n^2+n+1\right)\)

\(\Rightarrow n^{1986}-1⋮n^2+n+1\)

Mà \(\left\{{}\begin{matrix}n^{1988}-n^2⋮n^{1986}-1\\n^{1887}-n⋮n^{1986}-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n^{1988}-n^2⋮n^2+n+1\\n^{1987}-n⋮n^2+n+1\end{matrix}\right.\)

\(\Rightarrow B⋮n^2+n+1\)

Mà \(n^2+n+1>3\forall n>1\)

=> B ko là SNT với n > 1

Vậy n = 1 (T/m)

Phải là x + y =6 nhé bn, x + y = 4 ko xảy ra dấu =

\(\Rightarrow2A=6x+4y+\frac{12}{x}+\frac{16}{y}\)

\(=3\left(x+y\right)+\left(3x+\frac{12}{x}\right)+\left(y+\frac{16}{y}\right)\)

AD BDT Cô-si cho 2 số không âm

\(\Rightarrow3x+\frac{12}{x}\ge2\sqrt{36}=12;y+\frac{16}{y}\ge2\sqrt{16}=8\)

\(\Rightarrow2A\ge3.6+12+8=38\)

\(\Rightarrow A\ge19\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Vậy \(A_{min}=19\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

\(A=\frac{9x}{2-x}+\frac{2}{x}\)

\(=\frac{9x}{2-x}+\frac{2-x}{x}+1\)

AD BĐT Cosi cho 2 số thực không âm ta có:

\(\frac{9x}{2-x}+\frac{2-x}{x}\ge2\sqrt{\frac{9x}{2-x}.\frac{2-x}{x}}=2\sqrt{9}=6\)

\(\Rightarrow A\ge6+1=7\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{9x}{2-x}=\frac{2-x}{x}\Leftrightarrow x=\frac{1}{2}\)

Vậy \(A_{min}=7\Leftrightarrow x=\frac{1}{2}\)

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(a^3-c^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left[\left(a^3-b^3\right)+\left(b^3-c^3\right)\right]+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)

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

\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)

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

\(=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

Đặt \(x^2-x+3=a\) ta có:

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

\(=a\left(a-5\right)+4\)

\(=a^2-5a+4\)

\(=\left(a-1\right)\left(a-4\right)\)

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

\(a)DK:z\ne1\)

\(\left\{{}\begin{matrix}\frac{4}{z-1}+2x=7\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{z-1}+x=\frac{7}{2}=3,5\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-5y=-5\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=-8\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\5x=15\\\frac{2}{z-1}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\left(T/m\right)\)

Vậy ...

\(b)DK:\left\{{}\begin{matrix}x,y,z\ne0\\x,y,z>0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+\frac{1}{y}=2\\y+\frac{1}{z}=2\\z+\frac{1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}=6\)

\(\Leftrightarrow\left(x-2.\sqrt{x}.\frac{1}{\sqrt{x}}+\frac{1}{x}\right)+\left(y-2.\sqrt{y}.\frac{1}{\sqrt{y}}+\frac{1}{y}\right)+\left(z-2\sqrt{z}.\frac{1}{\sqrt{z}}+\frac{1}{z}\right)+2+2+2=6\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=0\)

Vì \(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2;\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2;\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=\frac{1}{\sqrt{x}}\\\sqrt{y}=\frac{1}{\sqrt{y}}\\\sqrt{z}=\frac{1}{\sqrt{z}}\end{matrix}\right.\)

\(\Leftrightarrow x=y=z=1\left(T/m\right)\)

Vậy ...

\(A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2010\)

\(=x^2-2x\left(y-z+1\right)+\left(y-z+1\right)^2+y^2+2z^2-4y+2yz-6z+2009\)

\(=\left[x-\left(y-z+1\right)\right]^2+y^2-2y\left(2-z\right)+\left(2-z\right)^2-\left(2-z\right)^2+2z^2-6z+2009\)

\(=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+z^2-2z+2005\)

\(=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+\left(z-1\right)^2+2004\ge2004\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y+z-1=0\\y-2+z=0\\z-1=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z=1\)

Vậy \(B_{min}=2004\Leftrightarrow x=y=z=1\)

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\Rightarrow\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(1+\sqrt{3}\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{9-3}\)

\(=\frac{6}{6}=1\)

\(\Rightarrow A=\sqrt{2}\)