a) \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\Rightarrow a+b+c=2;a,b,c>0\)

\(\Rightarrow S=\sqrt{\dfrac{\dfrac{ab}{2}}{\dfrac{ab}{2}+c}}+\sqrt{\dfrac{\dfrac{bc}{2}}{\dfrac{bc}{2}+a}}+\sqrt{\dfrac{ca}{ca+2b}}\)

\(=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)

Vì a,b,c>0 nên áp dụng BĐT AM-GM, ta có: 

 \(\sqrt{\dfrac{ab}{ab+2c}}=\sqrt{\dfrac{ab}{ab+\left(a+b+c\right)c}}=\sqrt{\dfrac{ab}{c^2+bc+ca+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)

\(=\sqrt{\dfrac{a}{a+c}}.\sqrt{\dfrac{b}{b+c}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\) 

\(\sqrt{\dfrac{bc}{bc+2a}}=\sqrt{\dfrac{bc}{\left(b+a\right)\left(c+a\right)}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)

\(\sqrt{\dfrac{ca}{ca+2b}}=\sqrt{\dfrac{ca}{\left(c+b\right)\left(a+b\right)}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)

\(\Rightarrow S\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi và chỉ khi: a=b=c=2/3=>\(\left(x,y,z\right)=\left\{\dfrac{2}{3};\dfrac{1}{3};\dfrac{2}{9}\right\}\)

Đặt \(\left(x;2y;3z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2\)

\(S=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)

\(S=\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}+\sqrt{\dfrac{bc}{bc+a\left(a+b+c\right)}}+\sqrt{\dfrac{ca}{ca+b\left(a+b+c\right)}}\)

\(S=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)

\(S\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\Rightarrow x;y;z\)

a: \(=-\dfrac{2}{a}\cdot x^2\cdot x^3\cdot y^3\cdot y\cdot z^2=-\dfrac{2}{a}x^5y^4z^2\)

b: \(=-a\cdot\dfrac{1}{4}\cdot\left(-b\right)^3\cdot x\cdot xy^3\cdot y^3=\dfrac{1}{4}ab^3x^2y^6\)

a, \(=\dfrac{-2x^5y^3z^2}{a}\)

b, \(=-\dfrac{xa\left(xy^3\right).1\left(-b^3y^3\right)}{4}=\dfrac{xa\left(b^3xy^6\right)}{4}=\dfrac{x^2ab^3y^6}{4}\)

\(2=4\sqrt{xy}+2\sqrt{xz}\le2x+2y+x+z=3x+2y+z\)

Ta có:

\(VT=\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}=2\left(\dfrac{xy}{z}+\dfrac{zx}{y}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{xy}{z}\right)+2\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\)

\(VT\ge2\left(x+y+z\right)+2y+4x\)

\(VT\ge2\left(3x+2y+z\right)\ge4\)

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

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

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

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

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

\(=\frac{-x-2-x+9-x+9}{1-x}=\frac{-3x+16}{1-x}\)

Câu b,c mk chưa học, bn thông cảm

Còn câu a, nếu sai thì xin lượng thứ :))

ĐK: \(x\ne1\)

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

\(=\frac{x+2}{x-1}-\frac{2\left(x-9\right)}{1-x}=\frac{x+2}{x-1}-\frac{2x-18}{1-x}\)

\(=\frac{x+2}{x-1}-\frac{\left(-1\right)\left(2x-18\right)}{\left(-1\right)\left(1-x\right)}=\frac{x+2}{x-1}-\frac{18-2x}{x-1}\)

\(=\frac{x+2-18+2x}{x-1}=\frac{3x-16}{x-1}\)

Bạn tham khảo:

cho x,y,z >0 thỏa mãn \(2\sqrt{y}+\sqrt{z}=\dfrac{1}{\sqrt{x}}\). CMR: \(\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}\ge... - Hoc24

Lời giải:
a)

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

\(=\frac{x+2}{x-1}+\frac{2(x-9)}{x-1}=\frac{x+2+2(x-9)}{x-1}=\frac{3x-16}{x-1}\)

b)

\(\frac{x^2-9y^2}{x^2y}: \frac{xz-3yz}{3xy}=\frac{x^2-9y^2}{x^2y}.\frac{3xy}{xz-3yz}\)

\(=\frac{(x-3y)(x+3y)}{x^2y}.\frac{3xy}{z(x-3y)}=\frac{3(x+3y)}{xz}\)

c) \(\frac{4(x+3)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4(x+3)}{3x-1}.\frac{3x-1}{x^2+3x}=\frac{4(x+3)}{x^2+3x}=\frac{4(x+3)}{x(x+3)}=\frac{4}{x}\)