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đặt \(\frac{a}{b}\)= \(\frac{c}{d}=k\Rightarrow\hept{\begin{cases}k=ab\\k=cd\end{cases}}\)
ta có : \(\frac{7a-4b}{3a+5b}\)= \(\frac{7ak-4b}{3ak-5b}=\frac{a\left(7k-4\right)}{a\left(3k-5\right)}=\frac{7k-4}{3k-5}\left(1\right)\)
\(\frac{7c-4d}{3c+5d}\)=\(\frac{7ck-4d}{3ck+5d}\)= \(\frac{c\left(7k-4\right)}{c\left(3k+5\right)}\)= \(\frac{7k-4}{3k+5}\)( 2 )
từ (1) và ( 2) => \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)( điều phải chứng minh )
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Giải:
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\left(=\frac{a}{c}\right)\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(đpcm\right)\)
Vậy...
Ta đặt:\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{7k-4}{3k+5}\)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{7k-4}{3k+5}\)
Do đó: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
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: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)