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Vì\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}\) = k
=> a = ck , b = dk
Thay a = ck , b = dk vào \(\frac{7a-11b}{4a+5b}\)ta có :
\(\frac{7a-11b}{4a+5b}=\frac{7.ck-11dk}{4ck+5dk}=\frac{k\left(7c-11d\right)}{k\left(4c+5d\right)}=\frac{7c-11d}{4c+5d}\)
Vậy \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\left(2a+3b\right)\left(4c-5d\right)=\left(4a-5b\right)\left(2c+3d\right)\)
\(\Leftrightarrow8ac-10ad+12bc-15bd=8ac+12ad-10bc-15bd\)
\(\Leftrightarrow-10ad+12bc=12ad-10bc\)
\(\Leftrightarrow\left(-10ad+12bc\right)+\left(-12bc-12ad\right)=\left(12ad-10bc\right)+\left(-12bc-12ad\right)\)
\(\Leftrightarrow22bc=22ad\)
a/b=c/d <=>a/c=b/d
=>4a/4c=5b/5d
Áp dụng.. ta có:
4a/4c=5b/5d=4a-5b/4c-5d=4a+5b/4c+5d
=>4a-5b/4a+5b=4c-5d/4c+5d(đpcm)
Tick nhé
Vì : a/b=c/d nên =>a/c=b/d
Đặt: a/c=b/d=k thì =>a=ck;b=dk
Thay :a=ck và b=dk vào 2a-3b/4a+5b có :
2a-3b/4a+5b=2ck-3dk/4ck+5dk=k(2c-3d)/k(4c+5d)=2c-3d/4c+5d
Tu đây suy ra : 2a-3b/4a+5b=2c-3d/4c+5d
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Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{4a}{4c}=\frac{5b}{5d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{4a}{4c}=\frac{5b}{5c}=\frac{4a-5b}{4c-5d}\) (1)
\(\frac{4a}{4c}=\frac{5b}{5d}=\frac{4a+5b}{4c+5d}\) (2)
Từ (1) và (2) => \(\frac{4a-5b}{4c-5d}=\frac{4a+5b}{4c+5d}\)
\(\Rightarrow\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\left(đpcm\right).\)
Chúc bạn học tốt!