\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5zx=6\left(z+x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{z+x}{zx}=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\b+c=\dfrac{4}{3}\\a+c=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\\c=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\\z=3\end{matrix}\right.\)

Vậy . . .

\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)

Đặt \(\dfrac{x}{y}=a;\dfrac{x}{y+12}=b;\dfrac{x}{y-12}=c\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\c-a=2\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(\dfrac{1}{b}-\dfrac{1}{a}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)

\(\Leftrightarrow\dfrac{a-b}{ab}+\dfrac{a-c}{ac}=0\Leftrightarrow\dfrac{1}{ab}-\dfrac{2}{ac}=0\)

\(\Leftrightarrow\dfrac{1}{a}\left(\dfrac{1}{b}-\dfrac{2}{c}\right)=0\Rightarrow\dfrac{1}{b}=\dfrac{2}{c}\Rightarrow c=2b\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\2b-a=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=4\\\dfrac{x}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4y\\\dfrac{4y}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=144\\y=36\end{matrix}\right.\)

Vậy . . .

\(\left\{{}\begin{matrix}x^2+y^2+x+y=8\left(1\right)\\x^2+y^2+xy=7\left(2\right)\end{matrix}\right.\)

(1) và (2), trừ vế theo vế

\(\Rightarrow x+y-xy=1\)

\(\Leftrightarrow\left(x-1\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(1-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Đến đây bn tự biện luận tiếp nhé (^ . ^). . .

\(M=x^2+xy+y^2-3x-3y\)

\(\Rightarrow4M=4x^2+4xy+4y^2-12x-12y\)

\(=\left(x^2+4y^2+9+4xy-12y-6x\right)+\left(3x^2-6x+3\right)-12\)

\(=\left(x+2y-3\right)^2+3\left(x-1\right)^2-12\ge-12\)

\(\Rightarrow M\ge-3\)

\(\Rightarrow Min_M=-3\Leftrightarrow x=y=1\)

\(M=\left(x+3\right)^2+\left(x-1\right)^2+2008\)

\(=\left(x^2+6x+9\right)+\left(x^2-2x+1\right)+2008\)

\(=\left(2x^2+4x+2\right)+2016\)

\(=2\left(x+1\right)^2+2016\ge2016\)

\(\Rightarrow Min_M=2016\Leftrightarrow x=-1\)

\(n=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)

\(\Rightarrow n^2=\left(3+\sqrt{5+2\sqrt{3}}\right)+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}+\left(3-\sqrt{5+2\sqrt{3}}\right)\)

\(=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}\)

\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=6+2\sqrt{3}-2=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow n=\sqrt{3}+1\)

\(\Rightarrow n^2-2n-2=\left(4+2\sqrt{3}\right)-2\left(\sqrt{3}+1\right)-2=0\)

\(\sqrt{4x-1}+\sqrt{4x^2-1}=1\)

\(\Leftrightarrow\sqrt{4x-1}-1+\sqrt{4x^2-1}=0\)

\(\Leftrightarrow\dfrac{4x-1-1}{\sqrt{4x-1}+1}+\sqrt{4x^2-1}=0\)

\(\Leftrightarrow\dfrac{2\left(2x-1\right)}{\sqrt{4x-1}+1}+\sqrt{\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\left(\dfrac{2\sqrt{2x-1}}{\sqrt{4x-1}+1}+\sqrt{2x+1}\right)\sqrt{2x-1}=0\)

\(\Rightarrow x=\dfrac{1}{2}\) (n)

Vậy phương trình có 1 nghiệm duy nhất . . .

\(\sqrt{x^2-4x+3}-\sqrt{2x^2-3x+1}=x-1\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-3\right)}-\sqrt{\left(x-1\right)\left(2x-1\right)}-\left(x-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-3}-\sqrt{2x-1}-\sqrt{x-1}\right)\sqrt{x-1}=0\)

Trường hợp 1:

\(\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(n\right)\)

Trường hợp 2:

\(\sqrt{x-3}-\sqrt{2x-1}-\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-3}-\sqrt{x-1}=\sqrt{2x-1}\)

\(\Leftrightarrow x-3-2\sqrt{x^2-4x+3}+x-1=2x-1\)

\(\Leftrightarrow2\sqrt{x^2-4x+3}=-3\) (vô lý)

Vậy phương trình có 1 nghiệm duy nhất . . .

Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b\ge0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\\left(1+b\right)^2+b^2-5=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2b^2+2b-4=0\)

\(\Leftrightarrow2\left(b-1\right)\left(b+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2+x-x^2}=1\)

\(\Leftrightarrow1=2+x-x^2\)

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\left(n\right)\)

Vậy phương trình có 2 nghiệm phân biệt . . .

Câu 3:

Áp dụng bđt AM - GM, ta có:

\(M=xyz\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

\(\le\dfrac{\left(x+y+z\right)^3}{27}\times\dfrac{\left[\left(x+y\right)+\left(x+z\right)+\left(y+z\right)\right]^3}{27}\)

\(=\dfrac{1^3}{27}\times\dfrac{2^3}{27}=\dfrac{8}{729}\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Câu 6:

Áp dụng bđt Cauchy Shwarz dạng Engel và bđt AM - GM, ta có:

\(M=\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\)

\(=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ac}+\dfrac{d^2}{ad+bd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{ad+bc+cd+ab+2ac+2bd}\)

\(=\dfrac{2\left(a+b+c+d\right)^2}{\left(2ad+2bc+2cd+2ab+2ac+2bd\right)+2ac+2bd}\)

\(\ge\dfrac{2\left(a+b+c+d\right)^2}{\left(2ad+2bc+2cd+2ab+2ac+2bd\right)+a^2+c^2+b^2+d^2}\)

\(=\dfrac{2\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=2\)

Dấu "=" xảy ra khi a = b = c = d