mn giúp em vs ạ em cần gấp cảm ơn ạ 

Bài 3: 

Gọi số học sinh là x

Theo đề, ta có: \(x\in BC\left(12;18;21\right)\)

hay x=504

CÂU 1: 

\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

CÂU 2: 

\(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)

CÂU 3: 

\(\dfrac{15x\left(x+5\right)^3}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)^2}{4x}\)

CÂU 4: 

\(\dfrac{3xy+x}{9y+3}=\dfrac{x\left(3y+1\right)}{3\left(3y+1\right)}=\dfrac{x}{3}\)

CÂU 5: 

\(\dfrac{3xy+3x}{9y+9}=\dfrac{3x\left(y+1\right)}{9\left(y+1\right)}=\dfrac{x}{3}\)

CÂU 6: 

\(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)

CÂU 7:

\(\dfrac{2x^2+2x}{x+1}=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

CÂU 8: 

\(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

CÂU 9: 

\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\dfrac{2y}{3\left(x+y\right)^2}\)

c) Ta có: \(\sqrt{\sqrt{x}+3}=3\)

\(\Leftrightarrow\sqrt{x}+3=9\)

\(\Leftrightarrow\sqrt{x}=6\)

hay x=36

Ta có: \(\sqrt{x-2\sqrt{x-1}}=2\)

\(\Leftrightarrow x-2\sqrt{x-1}-4=0\)

\(\Leftrightarrow x-1-2\cdot\sqrt{x-1}\cdot1+1=4\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=4\)

\(\Leftrightarrow\sqrt{x-1}-1=2\)

\(\Leftrightarrow\sqrt{x-1}=3\)

\(\Leftrightarrow x-1=9\)

hay x=10