Ta có : \(7x=3x\)

\(\frac{x}{3}=\frac{y}{7}\)

Áp dụng dãy tỉ số bằng nhau :

 \(\frac{x}{3}=\frac{y}{7}=\frac{2.x^2+y^2}{2.3^2+7^2}=\frac{268}{674}=4\)

\(\Rightarrow x=4.3=12\)

\(\Rightarrow y=4.7=28\)

Vậy x = 12 và y = 28

Help me!!!

Thanks

1. 2x = 3y-2

2x+2x = 3y

4x = 3y

=> \(\frac{x}{3}=\frac{y}{y}\Rightarrow\frac{x+y}{3+4}=\frac{14}{7}=2\)

=> \(\frac{x}{3}=2\Rightarrow x=6\)

=> \(\frac{y}{4}=2\Rightarrow y=8\)

hờ hờ vừa làm bài vừa mở olm

2.6X=5Y=X/5=Y/6=X-Y/5-6=10/-1=-10

X=-50;Y=-60

khó + lười + nhiều = không làm

Hello

ko thích làm

\(A=\left(2x\right)^2-2.2x.5+5^2-4x.x+4x.6\)

\(=4x^2-20x+25-4x^2+24x=4x+25\)

\(B=\left(7x-3y\right)^2-\left(7x-3y\right)\left(7x+3y\right)\)

\(=\left(7x-3y\right)\left(7x-3y-7x-3y\right)\)

\(=\left(7x-3y\right)\left(-6y\right)=18y^2-42xy\)

\(C=\left(3-2x\right)^2+\left(3+2x\right)^2\)

\(=9-2.3.2x+4x^2+9+2.3.2x+4x^2\)

\(=18+8x^2\)

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

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

a) \(7x=3y\Rightarrow\)\(\frac{x}{3}=\frac{y}{7}\)

Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow x=3k;y=7k\)

Có: x.y=84

\(\Rightarrow3k\cdot7k=84\)

\(\Rightarrow k^2=4\Rightarrow\left[\begin{array}{nghiempt}k=2\\k=-2\end{array}\right.\)

Với k=2 thì x=6 ;y=14

Với k=-2 thì x=-6 ;y =-14

b) \(7x=3y\Rightarrow\)\(\frac{x}{3}=\frac{y}{7}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}=\frac{y}{7}=\frac{5y-2x}{5\cdot7-2\cdot3}=\frac{-4}{29}\)

=> \(\begin{cases}x=-\frac{12}{29}\\y=-\frac{28}{29}\end{cases}\)

c) \(2x=3y=5z\)

\(\Leftrightarrow\)\(\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

Áp dụng tc của dãy tỉ số bằng nhau ta co:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+2y-3z}{15+2\cdot10-3\cdot6}\)

thiếu đề

a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)