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Ta đặt : a/b = c/d = K ( K khác 0 )
=> a = b.K
c = d.K
Mà : a2 + b2 / c2 + d2 = b.K2 + b2 / d.K2 + d2
= b2 . ( K2 + 1 ) / d2 . ( K2 + 1 )
= b2 / d2 ( 1 )
Mà : ab/cd = b.K.b / d.K.d = b2 . K / d2 . K
= b2 / d2 ( 2 )
Từ ( 1 ) và ( 2 ) suy ra : a/b =c/d ( ĐPCM )
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)
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.
1. Ta có: \(\dfrac{a}{b}=\dfrac{ab}{cd},\dfrac{c}{d}=\dfrac{bc}{bd}\)
a) Mẫu chung bd > 0 ( do b > 0, d > 0 ) nên nếu \(\dfrac{ad}{bd}< \dfrac{bc}{bd}\) thì ad < bc
b) Ngược lại, Nếu ad < bc thì \(\dfrac{ad}{bd}< \dfrac{bc}{bd}.\Rightarrow\dfrac{a}{b}< \dfrac{c}{d}\)
Ta có thể viết: \(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow ad< bc\)
2. a) Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) ( 1 )
Thêm ab vào 2 vế của (1): \(ad+ab< bc+ab\)
\(a\left(b+d\right)< b\left(a+c\right)\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) ( 2 )
Thêm cd vào 2 vế của (1): \(ad+cd< bc+cd\)
\(d\left(a+c\right)< c\left(b+d\right)\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) ( 3 )
Từ (2) và (3) ta có: \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
a) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\Rightarrow ad=cb\)
=> \(ad+bd=bc+bd\)
\(\Rightarrow d\left(a+b\right)=b\left(c+d\right)\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
\(\Rightarrow\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\rightarrowđpcm\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\rightarrowđpcm\)
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 \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a, Ta có: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)
\(\Rightarrow\dfrac{b^2.k}{d^2.k}=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b, Ta có:\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
CHÚC BẠN HỌC TỐT!!
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)( áp dụng tỉ lệ thức )
Ta đặt:
\(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2.\left(c.d\right)}{c.d}=k^2\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(ck+dk\right)^2}{\left(c+d\right)^2}=\dfrac{k^2.\left(c+d\right)^2}{\left(c+d\right)^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(ck\right)^2+\left(dk\right)^2}{c^2+d^2}=\dfrac{c^2k^2+d^2k^2}{c^2+d^2}=\dfrac{k^2.\left(c^2+d^2\right)}{c^2+d^2}=k^2\) (3)
Từ (1) và (3) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
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 \(\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có: \(\frac{a}{b}=\frac{b}{d}\Rightarrow ad=b^2\)
Thay \(ad=b^2\), ta có
\(\frac{a^2+b^2}{b^2+d^2}=\frac{a^2+ad}{+ad+d^2}=\frac{\left(a+d\right)a}{\left(a+d\right)d}=\frac{a}{d}\)
Vậy\(\frac{a^2+b^2}{b^2+d^2}=\frac{a}{d}\)khi\(\frac{a}{b}=\frac{b}{d}\)