\(\frac{2}{3}x\frac{4}{y}=\frac{4}{45}:\frac{1}{5}\)\(=\frac{4}{15}\)

\(\frac{4}{y}=\frac{4}{15}:\frac{2}{3}\)\(=\frac{2}{5}\)

y=4:\(\frac{2}{5}\)    y=10

\(\frac{3}{4}x\frac{y}{5}=\frac{15}{4}\)

\(\frac{y}{5}=\frac{15}{4}:\frac{3}{4}=5\)

y=5x5=25

Hệ đã cho \(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-4\left(x+y\right)+4=49\\\left(x-y\right)^2-2\left(x-y\right)+1=4\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+y-2\right)^2=49\\\left(x-y-1\right)^2=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=9\\x-y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=3\end{cases}}}\)

Vậy nghiệm của hệ là (x;y) = (6;3)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\) 

\(\frac{x}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\frac{x+y+z}{4+5+6}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\) mà x + y + z = 45

\(\Rightarrow\frac{45}{15}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow3=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\hept{\begin{cases}x=3\cdot4=12\\y=3\cdot6=18\\z=3\cdot5=15\end{cases}}\)

Noob vl

Ta có : Vì \(\left(x+y\right)^2-4\left(x+y\right)=45\)

\(\Rightarrow\left(x+y\right)^2-2.\left(x+y\right).2+2^2=49\)

\(\Rightarrow\left(x+y-2\right)^2=49\)\(\Rightarrow\orbr{\begin{cases}x+y-2=7\\x+y-2=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x+y=9\\x+y=-5\end{cases}}\)

Vì \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(x-y\right)^2-2.\left(x-y\right)+1^2=4\)

\(\Rightarrow\left(x-y+1\right)^2=4\)

\(\Rightarrow\orbr{\begin{cases}x-y-1=2\\x-y-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x-y=3\\x-y=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=y+3\\x=y-1\end{cases}}\)

* \(x+y=9\)\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=3;x=6\\x=y-1\Rightarrow y=5;x=4\end{cases}}\)

* \(x+y=-5\)\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=-4;x=-1\\x=y-1\Rightarrow y=-2;x=-3\end{cases}}\)

Vậy cặp số x;y là (6;3) , (4;5) , (-1;-4) , (-3;-2) 

Kb với tớ nhé, mn!

yêu cầu là j ạ

Ta có : Vì \left(x+y\right)^2-4\left(x+y\right)=45(x+y)2−4(x+y)=45

\Rightarrow\left(x+y\right)^2-2.\left(x+y\right).2+2^2=49⇒(x+y)2−2.(x+y).2+22=49

\Rightarrow\left(x+y-2\right)^2=49⇒(x+y−2)2=49\(\Rightarrow\orbr{\begin{cases}x+y-2=7\\x+y-2=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x+y=9\\x+y=-5\end{cases}}\)

Vì \left(x-y\right)^2-2\left(x-y\right)=3(x−y)2−2(x−y)=3

\Rightarrow\left(x-y\right)^2-2.\left(x-y\right)+1^2=4⇒(x−y)2−2.(x−y)+12=4

\Rightarrow\left(x-y+1\right)^2=4⇒(x−y+1)2=4

\(\Rightarrow\orbr{\begin{cases}x-y-1=2\\x-y-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x-y=3\\x-y=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=y+3\\x=y-1\end{cases}}\)

* x+y=9x+y=9\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=3;x=6\\x=y-1\Rightarrow y=5;x=4\end{cases}}\)

* x+y=-5x+y=−5\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=-4;x=-1\\x=y-1\Rightarrow y=-2;x=-3\end{cases}}\)

Vậy cặp số x;y là (6;3) , (4;5) , (-1;-4) , (-3;-2) 

a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)

a: Ta có: \(x^2-2xy+y^2-4z^2\)

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

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

\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)

\(=-8000\)

b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)

\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)

\(=3x^2+12x-63+x^2-8x+16+48\)

\(=2x^2+4x+1\)

\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)

\(=\dfrac{7}{2}\)

Áp dụng BĐT AM-GM ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\)

Suy ra: \(P=6\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+8\left[\left(x^2+y^2\right)^2-2\left(xy\right)^2\right]+\frac{5}{xy}\)

\(\ge6\left(1-\frac{3}{4}\right)+8\left(\frac{1}{4}-\frac{1}{8}\right)+\frac{5}{\frac{1}{4}}\) (Do x+y=1) \(\Rightarrow P\ge6-\frac{9}{2}+2-1+20=\frac{45}{2}\)(đpcm).

Dấu "=" xảy ra <=> x=y=1/2.