Đặt \(\frac{a}{b}\) = \(\frac{c}{d}\) = k ⇒ \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\frac{ab}{cd}\) = \(\frac{bk.b}{dk.d}\) = \(\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) = \(\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\) = \(\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}\) = \(\frac{b^2}{d^2}\)
⇒ \(\frac{ab}{cd}\) = \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
cho a^2+b^2/c^2+d^2=ab/cd. CMR a/b=c/d hoặc a/b=d/c
cho a^2+b^2/c^2+d^2=ab/cd cmr a/b=c/d hoặc a/b = d/c
cho a/b = c/d, cmr: \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Cho \(\dfrac{a}{b}< \dfrac{c}{d}\)và b, d > 0 . CMR : \(\dfrac{a}{b}< \dfrac{ab+cd}{b^2+d^2}< \dfrac{c}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\)
a, \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b, \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
Cho \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
CMR \(\left[{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)
cmr \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
Biết \(\dfrac{a}{b}< \dfrac{c}{d}\left(b,d>0\right)\)
CMR \(\dfrac{a}{b}=\dfrac{ab+cd}{b^2+d^2}< \dfrac{c}{d}\)
