a) Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{\left(bk\right)^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\) (1)
Tương tự, ta cũng có \(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(k+1\right)^2}{k^2+1}\) (2)
Từ (1), (2) suy ra \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\)
a) Ta có: ab=cd⇒ac=bdab=cd⇒ac=bd
⇒a2c2=b2d2=abcd⇒a2c2=b2d2=abcd
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
a2c2=b2d2=a2−b2c2−d2a2c2=b2d2=a2−b2c2−d2
⇒a2−b2c2−d2=abcd⇒a2−b2c2−d2=abcd
⇒a2−b2ab=c2−d2cd
b) Đặt ab=cd=k⇒{a=bkc=dkab=cd=k⇒{a=bkc=dk
Ta có:
(a+b)2a2+b2=(bk+b)2(bk)2+b2=b2(k+1)2b2(k2+1)=(k+1)2k2+1(a+b)2a2+b2=(bk+b)2(bk)2+b2=b2(k+1)2b2(k2+1)=(k+1)2k2+1 (1)
Tương tự, ta cũng có (c+d)2c2+d2=(k+1)2k2+1(c+d)2c2+d2=(k+1)2k2+1 (2)
Từ (1), (2) suy ra (a+b)2a2+b2=(c+d)2c2+d2
