Lớp 9 không biết có học tới sin cos âm chưa nếu chưa thì lấy phần dương nha 

\(1+tan^2a=\frac{1}{cos^2a}\)    

\(1+\left(\frac{2}{3}\right)^2=\frac{1}{cos^2a}\)    

\(1+\frac{4}{9}=\frac{1}{cos^2a}\)    

\(\frac{13}{9}=\frac{1}{cos^2a}\)    

\(cos^2a=\frac{9}{13}\)   

\(cosa=\pm\sqrt{\frac{9}{13}}=\pm\frac{3\sqrt{13}}{13}\)    

\(sin^2a+cos^2a=1\)   

\(sin^2a+\frac{9}{13}=1\)    

\(sin^2a=\frac{4}{13}\)    

\(sina=\pm\sqrt{\frac{4}{13}}=\pm\frac{2\sqrt{13}}{13}\)   

tan dương nên sẽ có 2 TH 

TH1 sin và cos cùng dương 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{2\sqrt{13}}{13}\right)^3+3\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}\)    

\(=-\frac{89}{459}\)   

TH2 sin và cos cùng âm 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{-2\sqrt{13}}{13}\right)^3+\left(\frac{-3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{-2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{-3\sqrt{13}}{13}\right)^3}\)

\(=-\frac{89}{459}\)

VT=\(\frac{a^2}{ab+\frac{1}{b}}+\frac{b^2}{bc+\frac{1}{c}}+\frac{c^2}{ca+\frac{1}{a}}\)

áp dụng bđt cộng mẫu đc VT \(\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left(a+b+c\right)^2}{ab+bc+ca+\frac{ab+bc+ca}{abc}}\left(1\right)\)

Ta có  \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\forall a,b,c\)

Nên \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+\frac{\left(a+b+c\right)^2}{3abc}}=\frac{1}{\frac{1}{3}+\frac{1}{3abc}}=\frac{3abc}{1+abc}\left(đccm\right)\)

dấu bằng xảy ra <> a=b=c

1. 

A = \(\frac{x^2-10y^2+10z^2}{y+z}+\frac{y^2-10z^2+10x^2}{z+x}+\frac{z^2-10x^2+10y^2}{x+y}\)

A = \(\frac{x^2}{y+z}-10\frac{\left(y-z\right)\left(y+z\right)}{y+z}+\frac{y^2}{x+z}-10\frac{\left(z-x\right)\left(z+x\right)}{z+x}+\frac{z^2}{x+y}-10\frac{\left(x-y\right)\left(x+y\right)}{x+y}\)

A = \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}-10y+10z-10z+10x-10x+10y\)

\(A=x\left(\frac{x}{y+z}+1\right)+y\left(\frac{y}{z+x}+1\right)+z\left(\frac{z}{x+y}+1\right)-\left(x+y+z\right)\)

A = \(x\cdot\frac{x+y+z}{y+z}+y\cdot\frac{x+y+z}{z+x}+y\cdot\frac{x+y+z}{x+y}-\left(x+y+z\right)\)

A = \(\left(x+y+z\right)\cdot\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)

A = \(\left(x+y+z\right)-\left(x+y+z\right)=0\)

Mn giải giúp e vs ((