Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,d=ck\)
a) \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk.b}=\dfrac{b^2\left(k^2-1\right)}{b^2.k}=\dfrac{k^2-1}{k}\) (1)
\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk.d}=\dfrac{d^2\left(k^2-1\right)}{d^2k}=\dfrac{k^2-1}{k}\) (2)
Tử (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b) \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{b^2\left(k^2+1\right)}\)
\(=\dfrac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (1)
\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left[d\left(k+1\right)\right]^2}{d^2\left(k^2+1\right)}\)
\(=\dfrac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
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