Sửa đề: Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh các tỉ lệ thức sau
a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{5a+3b}{5c+3d}=\frac{5\cdot bk+3b}{5\cdot dk+3d}=\frac{b\left(5k+3\right)}{d\left(5k+3\right)}=\frac{b}{d}\)
\(\frac{5a-3b}{5c-3d}=\frac{5\cdot bk-3b}{5\cdot dk-3d}=\frac{b\left(5k-3\right)}{d\left(5k-3\right)}=\frac{b}{d}\)
Do đó: \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
b: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
