Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow a=bk;c=dk\)
\(\Rightarrow\frac{4a-3b}{a}=\frac{4bk-3b}{bk}=\frac{b\left(4k-3\right)}{bk}=\frac{4k-3}{k}\left(1\right)\)
\(\Rightarrow\frac{4c-3d}{c}=\frac{4dk-3d}{dk}=\frac{d\left(4k-3\right)}{dk}=\frac{4k-3}{k}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrow\frac{4a-3b}{a}=\frac{4c-3d}{c}\left(đpcm\right)\)
Chứng minh \(\dfrac{a}{b} = \dfrac{c}{d}\) nếu biết
a, \(\dfrac {4a-3b}{4c-3d} = \dfrac {4a+3b}{4c+3d}\)
b, \(\dfrac {2a-3b}{2a+3b} = \dfrac {2c-3d}{2c+3d}\)
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \((a+c).((b-d)=(a-c).(b-d)\)
b, \((a+c).b=(b+d).a\)
c, \(a.(b-d)=b(a-c)\)
d, \((b+d).c=(a+c).d\)
e, \((b-d).c=(a-c).d\)
f, \((a+b).(c-d)=(a-b).(c+d)\)
g, \((2a+3c).(2b-3d)=(2a-3c).(2b+3d)\)
h, \((4a+3b).(4c-3d)=(4a-3b).((4c+3d)\)
i, \((2a+3b).(4c-5d)=(4a-5b).(2c+3d)\)
k, \((4a+5b).(7c-11d)=(7a-11b).(4c+5d)\)
Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . CMR :
\(a,\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
\(b,\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{a^3-b^3}{c^3-d^3}\)
Bài 1:
a) Cho tỉ lệ thức \(\frac{3x-y}{x+y}=\frac{3}{4}\) tính giá trị của \(\frac{x}{y}\)
b) Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Cho tỉ lệ thức \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng:
a)\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
b)\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
Chứng minh \(\dfrac {a}{b} = \dfrac {c}{d}\) nếu biết
a, \(\dfrac {4a-3b}{a}=\dfrac {4c-3d}{c}\)
b, \(\dfrac {1111c-99d}{9999-11d} = \dfrac {1111a-99b}{9999a-11b}\)
cho tỉ lệ thức:
\(\dfrac{7a+3b}{7a-3b}=\dfrac{7c+3d}{7c-3d}\).CMR: \(\dfrac{a}{b}=\dfrac{c}{d}\)(giả sử các tỉ số đều có nghĩa)
