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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 được :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\) \(\left(1\right)\)
Tương tự :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Vậy \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) khi \(\dfrac{a}{b}=\dfrac{c}{d}\)
Đặt: a/b = c/d = k => a = bk, c = dk
Ta có:
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) => a+b/a-b = c+d/c-d
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.
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)
Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)
+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)
a: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
d: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\) suy ra \(\dfrac{a}{c}=\dfrac{b}{d}\)
Theo tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Suy ra: \(\dfrac{a+b}{a-c}=\dfrac{c+d}{c-d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk\) và \(c=dk\)
Nên \(\dfrac{a+b}{c-d}=\dfrac{bk+b}{dk-d}=\dfrac{b\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\Rightarrow\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) (với \(a-b\ne0,c-d\ne0\))
Vậy \(\dfrac{a}{b}=\dfrac{c}{d}thì\)\(\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) ( \(a-b\ne0,c-d\ne0\))
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có VT:
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)
\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)
VT: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\) (2)
Từ (1) và (2)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)
Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ab=cd\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)\(\Leftrightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
Vậy...
Bài giải:
Với \(a,b,c,d\ne0\) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{b}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\Rightarrow\dfrac{a-b}{c-d}=\dfrac{b}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(ĐPCM\right)\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Khi đó:
\(\dfrac{a+b}{a-b}=\dfrac{bt+b}{bt-b}=\dfrac{b\left(t+1\right)}{b\left(t-1\right)}=\dfrac{t+1}{t-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dt+d}{dt-d}=\dfrac{d\left(t+1\right)}{d\left(t-1\right)}=\dfrac{t+1}{t-1}\)
Ta có đpcm