mong mn lm giúp tôi

Theo đề, ta có: \(\frac{x}{2}=\frac{y}{-5}\)và \(x-y=-7\)

Theo TC dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{-7}{7}=-1\)

\(\Leftrightarrow\hept{\begin{cases}x=-1.2=-2\\y=-1.-5=5\end{cases}}\)

a.x=12 ,y=16

a) theo đề, ta có:\(\frac{x}{3}=\frac{y}{4}\)và \(x+y=28\)

Theo TC dãy tỉ số bằng nhau, ta có: 

\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{28}{7}=4\)

\(\Leftrightarrow\hept{\begin{cases}x=4.3=12\\y=4.4=16\end{cases}}\)

Ta có:

\(2P=\frac{2x^2}{y^2}+\frac{2y^2}{x^2}-6\left(\frac{x}{y}+\frac{y}{x}\right)+10\)

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

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

\(\ge2\)

\(\Rightarrow P\ge1\)

Dấu = xảy ra khi x = y

gì mà  đánh võng kính thế

\(\frac{x}{y}=\frac{2}{5}\Rightarrow\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{x^2+y^2}{4+25}=\frac{29}{29}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{4}=1\\\frac{y^2}{25}=1\end{cases}\Rightarrow\hept{\begin{cases}x^2=4\\y^2=25\end{cases}\Rightarrow}\hept{\begin{cases}x=\pm2\\y=\pm5\end{cases}}}\)

\(\left(x-\frac{1}{3}\right)\left(y-\frac{1}{2}\right)\left(z-5\right)=0.\)

\(\Rightarrow x-\frac{1}{3}=0\)hoặc \(y-\frac{1}{2}=0\)hoặc \(z-5=0\)

TH1: \(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\Rightarrow\hept{\begin{cases}y=\frac{4}{3}\\z=-\frac{2}{3}\end{cases}}\)

Xét 2 trường hợp cpnf lại ta được ba bộ số x,y,z cần tìm

x=6

y=  -15

\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=-\frac{21}{7}=-3\)

\(\Rightarrow\frac{x}{2}=-3\Rightarrow x=-3\cdot2=-6\)

\(\Rightarrow\frac{y}{5}=-3\Rightarrow y=-3.5=-15\)

-6 và -15 viết nhầm à linh

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{25}{2}\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)