Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh:
1. \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\) 2. \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
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DH
1
12 tháng 12 2021
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\end{matrix}\right.\\ \RightarrowĐpcm\)
17 tháng 6 2019
a)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{c}{d}\) =>\(\frac{a}{c}=\frac{b}{d}\)
=>\(\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
=>\(\frac{ac}{bd}=\frac{a^2+b^2}{c^2+d^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\Rightarrow\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2+b^2}{a^2-b^2}=\frac{c^2+d^2}{c^2-d^2}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
6 tháng 12 2017
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=\dfrac{b^2t}{d^2t}=\dfrac{b^2}{d^2}\\\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2t^2+b^2}{d^2t^2+d^2}=\dfrac{b^2\left(t^2+1\right)}{d^2\left(t^2+1\right)}=\dfrac{b^2}{d^2}\end{matrix}\right.\Rightarrowđpcm\)
b)\(\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{t^2bd}{bd}=t^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2t^2+d^2t^2}{b^2+d^2}=\dfrac{t^2\left(b^2+d^2\right)}{b^2+d^2}\end{matrix}\right.\Rightarrowđpcm\)
17 tháng 8 2017
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow a=bk;c=dk\)
\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
17 tháng 8 2017
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) =>\(a=bk,c=dk\)
=> \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)
=\(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1)và(2)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Chúc Bạn Học Tốt
5 tháng 11 2021
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}=k^2\)
\(\dfrac{ac}{bd}=k^2\)
Do đó: \(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
DN
1
2 tháng 11 2017
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\left(1\right)\)
Thay (1) vào từng vế của đề bài:
\(VT=\dfrac{a^2+ac}{c^2-ac}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
Vế phải đặt thừa số chung sẽ ra VT => đpcm.
10 tháng 11 2018
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
10 tháng 11 2018
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
11 tháng 12 2022
a; Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7b^2k^2-3\cdot bk\cdot b}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2-3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2-3k}{11k^2-8}\)
\(\dfrac{7c^2-3cd}{11c^2-8d^2}=\dfrac{7d^2k^2-3kd^2}{11d^2k^2-8d^2}=\dfrac{7k^2-3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2-3cd}{11c^2-8d^2}\)
NB
8 tháng 12 2021
Xét \(\left(a^2+b^2\right).C-\left(b^2+c^2\right).a=a^2c+b^2a\)=\(b^2a-c^2a=a^2c+ac.c-ac.a=0\)
(thay \(b^2=ac\))
\(\Rightarrow\left(a^2+b^2\right).c=\left(b^2+c^2\right).a\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
1: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
Do đó; \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
\(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{b^2k^2-d^2k^2}{b^2-d^2}=k^2\)
Do đó: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)