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Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
Cách 1 :
Từ a/b = c/d => a/c = b/d ( tính chất tỉ lệ thức )
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
a/c = b/d = a+b/a-b = a-b/c-d => a+b/a-b = c+d/c-d ( tính chất tỉ lệ thức )
Vậy a+b/a-b = c+d/c-d
Cách 2:
Đặt : a/b = c/d = k
a/b = k => a= bk
c/d = k => c=dk
a+b/a-b = bk+b/ bk-b = b(k+1)/b(k-1) = k+1/k-1. (1)
c+d/c-d = dk+d/dk-d = d(k+1)/d(k-1) + k+1/k-1. (2)
Từ (1) và (2) suy ra a+b/a-b = c+d/c-d.
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Vậy...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) (1)
Thay (1) vào:
\(\dfrac{a+b}{a-b}=\dfrac{b.k+b}{b.k-b}=\dfrac{b.\left(k+1\right)}{b.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
\(\dfrac{c+d}{c-d}=\dfrac{d.k+d}{d.k-d}=\dfrac{d.\left(k+1\right)}{d.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (3)
Từ (2) và (3) =>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}=\dfrac{k+1}{k-1}\)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\Rightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\\ ac+bc-2ad-2bd=ac+ad-2bc-2bd\\ bc-2ad=ad-2bc\\ 3bc=3ad\\ bc=ad\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
\(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\)
\(\Leftrightarrow\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)
\(\Leftrightarrow ac-2ad+bc-2bd=ac-2bc+ad-2bd\)
\(\Leftrightarrow2ad+ad=2bc+bc\)
\(\Leftrightarrow3ad=3bc\)
\(\Leftrightarrow ad=bc\rightarrowđpcm\)
a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)(đpcm)
b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{a}{c}-1=\dfrac{b}{d}-1\)
=>\(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)(đpcm)
a)\(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
=>\(\dfrac{a}{b}\) -1= \(\dfrac{c}{d}\) -1
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\\ \Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\\ \dfrac{a^2}{c^2}=\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Có thể dùng cách khác:v
a)\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=t\)(với t là 1 số thực bất kì thỏa mãn)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)
Tương tự:v
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk và c=dk
ta có \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)\(\dfrac{ab}{cd}=\dfrac{bk.b}{bk.d}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (cùng =\(\dfrac{b^2}{d^2}\) ) (đpcm)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=t\)
a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\left(\dfrac{a+b}{c+d}\right)^2=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)
a) Ta co: a/b = c/d= k
=> a=bk
c=dk
Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)
Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)
Tu (1) va (2)
=> a-b/a+b=c-d/c+d
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b) Từ (*) ta có:
\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)
Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)
\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)
Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
đặt a/b=c/d=k
=>a=bk;c=dk rồi cứ thế thay lần lượt vào ac/bd;a^2+c^2/b^2+d^2
full hd :))
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Khi đó:
\(\dfrac{ac}{bd}=\dfrac{bt.dt}{bd}=\dfrac{t^2bd}{bd}=t^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2t^2+d^2t^2}{b^2+d^2}=\dfrac{t^2\left(b^2+d^2\right)}{b^2+d^2}=t^2\)
Vậy.....
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
=> (a+b)(c-d) = (c+d)(a-b)
=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Cách 1 :
Ta có: (a+b)(c-d)=ac-ad+bc-bd (1)
(a-b)(c+d)=ac+ad-bc-bd (2)
Từ giả thiết: \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> ad=bc (3)
Từ (1),(2),(3)=> (a+b)(c-d)=(a-b)(c+d)=>\(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
C2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k=>a=bk,c=dk\)
Ta có: \(\dfrac{a+b}{a-b}=\dfrac{kb+b}{kb-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\) (1)
\(\dfrac{c+d}{c-d}=\dfrac{kd+d}{kd-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
Từ (1) và (2) =>: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
C3:
Từ giả thiết: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=>\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)