\(\text{Câu}1:\)
\(\text{Đặt}\) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) \(\left(1\right)\)
\(\text{Từ}\) \(\left(1\right)\) \(\text{suy ra:}\)
\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\text{Vậy}\) \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) \(\left(ĐPCM\right)\)
\(Câu2:\) \(\text{Giải}\)
\(\text{Gọi số học sinh của 4 khối 6;7;8;9 lần lượt là a;b;c;d }\)\(\left(a;b;c;d\in N\text{*}\right)\)
\(\text{Theo bài ra ta có :}\)
\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}\)
\(b-d=70\)
\(\text{ Áp dụng tính chất dãy tỉ số bằng nhau ta được:}\)
\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}=\dfrac{b-d}{8-6}=\dfrac{70}{2}=35\) \(\left(1\right)\)
\(\text{Từ}\) \(\left(1\right)\) \(\text{suy ra:}\)
\(\dfrac{a}{9}=35\Rightarrow a=315\)
\(\dfrac{b}{8}=35\Rightarrow b=280\)
\(\dfrac{c}{7}=35\Rightarrow c=245\)
\(\dfrac{d}{6}=35\Rightarrow d=210\)
\(\text{ Vậy}\) \(a=315\\ b=280\\ c=245\\ d=210\)
\(\)
