biết:\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d\(\ne\)0. CMR:
\(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
Biết \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) (a,b,c,d \(\ne\) 0). CMR : \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Ta có :
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
\(Cho\) : \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d ≠ 0;c ≠ d,-d
Chứng minh rằng : \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)(b\(\ne\)0;d\(\ne\)0)
c)\(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
d)\(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)
\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)
Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
cho\(\dfrac{a}{b}=\dfrac{c}{d}\)với c\(\ne\) \(\pm\)1. CMR \(\dfrac{ab}{cd}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
Biết \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d khác 0. Chứng minh rằng: \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
ta có \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\Rightarrow ab.\left(c^2+d^2\right)=cd.\left(a^2+b^2\right)\)
suy ra \(ab.\left(c^2+d^2\right)\)=\(abc^2+abd^2=acbc+adbd\) (1)
\(cd\left(a^2+b^2\right)=a^2cd+b^2cd+bcbd\) =acad+bcbd (2)
(1);(2) suy ra acbc+adbd=acad+bcbd
nên bc+ad=bc+ad
suy ra ad=bc nên \(\dfrac{a}{b}=\dfrac{c}{d}\)
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)với \(a,b,c,d\ne0\); \(c\ne\pm d\). Chứng minh rằng \(\dfrac{a}{b}=\dfrac{c}{d}\)hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)
a, cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (b,d \(\ne\)0) CMR:\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b,cho \(\dfrac{a}{b}=\dfrac{c}{d}\)(b,d \(\ne\)0) CMR:\(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)
Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
cho \(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{ab}{cd}\).Chứng minh rằng: hoặc \(\dfrac{a}{b}\)= \(\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}\)= \(\dfrac{d}{c}\)