Cho tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\)chứng tỏ rằng :\(\frac{3a^2+5b^2}{3c^2+5d^2}=\frac{2a^2+7ab}{2c^2+7cd}\)
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cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b) \(\frac{2a+3b}{2a-3b}\)=\(\frac{2c+3d}{2c-3d}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)
a,\(\frac{a-b}{a+b}=\frac{c-d}{c+d};\)
b,\(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d};\)
c,\(\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2};\)
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
c, \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
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cho tỉ lệ thức\(\frac{a}{b}=\frac{c}{d}\).chứng minh
a) \(\frac{3a+5b}{3a-5b}=\frac{3c+5b}{3c-5b}\)
b)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). Chứng minh: \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
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Cho tỉ lệ thức a/b=c/d.Chứng minh
a)3a+5b/3a-5b=3c+5d/3c-5d
b) 2a + 3b/ 2a - 3b= 2c+3d/2c-3d
c)ab/cd=a^2-b^2/c^2-d^2
cho tỉ lệ thức \(\frac{a}{b}\)= \(\frac{c}{d}\),cmr ta có tỉ lệ thức sau:
a) \(\frac{3a+5b}{3a-5b}\)= \(\frac{3c+5d}{3c-5d}\)
b) \(\frac{7a^2+5ac}{7a^2-5ac}\)=\(\frac{7b^2+5bd}{7b^2-5bd}\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) chứng minh \(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Ta có:
a/b=c/d => a/c=b/d=2a/2c=3b/3d
= 2a+3b/2c+3d=2a-3b/2c-3d
=> 2a+3b/2a-3b=2c+3d/2c-3d (ĐPCM)
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1. Cho tỉ lệ thức: a/b = c/d chứng tỏ rằng:
2a + 5b/3a -4b = 2c + 5d/3c - 4d
Ta đặt:\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)
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