Gọi số hs của khối 6,7,8,9 lần lượt là a,b,c,d, ta có;

\(\dfrac{a}{6}=\dfrac{b}{5}=\dfrac{c}{4}=\dfrac{d}{3}\)

Áp dụng t/c DTSBN:

\(\dfrac{a}{6}=\dfrac{c}{4}=\dfrac{a-c}{6-4}=\dfrac{84}{2}=42\)

\(\dfrac{a}{6}=42\Rightarrow a=252\)

\(\dfrac{b}{5}=42\Rightarrow b=210\)

\(\dfrac{c}{4}=42\Rightarrow c=168\)

\(\dfrac{d}{3}=42\Rightarrow d=126\)

Vậy...

\(a,\dfrac{4x+3}{9x}+\dfrac{5x-3}{9x}=\dfrac{4x+3+5x-3}{9x}=\dfrac{9x}{9x}=1\\ b,\dfrac{x^2}{x+y}+\dfrac{y^2}{x+y}+\dfrac{2xy}{x+y}=\dfrac{x^2+y^2+2xy}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)

Sai đề bạn nhé!

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-\left(x^2+5x\right)=0\\ 2\left(x+5\right)-\left[x\left(x+5\right)\right]=0\\ \left(2-x\right)\left(x+5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

\(4x^3+15=47\\ 4x^3=32\\ x^3=8\\ x=2\)

D

Chiều cao trung bình của mỗi bạn là:

\(\dfrac{138+132}{2}=135\left(cm\right)\)