cho:\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
chung min\(\dfrac{a}{b}=\dfrac{c}{d}\)hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
cho \(\dfrac{a^2+b^2}{c^2+d^2}\)= \(\dfrac{ab}{cd}\).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 \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
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 \(\dfrac{a}{b}=\dfrac{d}{c}\) ?
bài 2
Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Ta có :\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow cd\left(a^2+b^2\right)=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=c^2ab+d^2ab\)
\(\Leftrightarrow\left(a^2cd+b^2cd\right)-\left(c^2ab+d^2ab\right)=0\)
\(\Leftrightarrow aacd+bbcd-ccab-ddab=0\)(tất cả là dấu nhân ko phải số tự nhiên có 4 chữ số nha)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\left(\text{đ}pcm\right)\)
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 ≠ 0;c ≠ d,-d
Chứng minh rằng : \(\dfrac{a}{b}=\dfrac{c}{d}\) 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 \(a,b,c,d\ne0\); \(c\ne\pm d\). 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)
1. Cho tỉ lệ thức \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\). CMR:
a) \(\dfrac{3a+5c}{3b+5d}\) = \(\dfrac{a-2c}{b-2d}\).
b) \(\dfrac{a^2-b^2}{ab}\) = \(\dfrac{c^2-d^2}{cd}\).
c) \(\dfrac{\left(a+b\right)^2}{a^2+b^2}\) = \(\dfrac{\left(c+d\right)^2}{c^2+d^2}\).
d) \(\left(\dfrac{a+b}{c+d}\right)^3\) = \(\dfrac{a^3+b^3}{c^3+d^3}\).
Gíup mình với cảm ơn các bạn rất nhiều!!!!!!!!!
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)
\(\dfrac{a-2c}{b-2d}=\dfrac{bk-2dk}{b-2d}=\dfrac{k\left(b-2d\right)}{b-2d}=k\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)
b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)
\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\left(dpcm\right)\)
c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)
giúp mình câu d) luôn nha phong
cảm ơn phong nha
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 \(\dfrac{a}{b}=\dfrac{d}{c}\)
(a² + b²) / (c² + d²) = ab/cd
<=> (a² + b²)cd = ab(c² + d²)
<=> a²cd + b²cd = abc² + abd²
<=> a²cd - abc² - abd² + b²cd = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> ac - bd = 0 hoặc ad - bc = 0
<=> ac = bd hoặc ad = bc
<=> a/b = d/c hoặc a/b = c/d (đpcm)