Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
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
Ta có: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
mà \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
Chúc bạn học tốt!
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: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2-3cd}{11c^2-8d^2}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
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.
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k (1)
=> a=bk ,c=dk
a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)
b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
a)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{c}{d}\) =>\(\frac{a}{c}=\frac{b}{d}\)
=>\(\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
=>\(\frac{ac}{bd}=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\Rightarrow\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2+b^2}{a^2-b^2}=\frac{c^2+d^2}{c^2-d^2}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
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 \(b^2=ac\))
\(\Rightarrow\left(a^2+b^2\right).c=\left(b^2+c^2\right).a\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
cảm ơn
