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Xem ở Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
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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...
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Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{2a+b}{3a-5b}=\dfrac{2\cdot bk+b}{3\cdot bk-5b}=\dfrac{2k+1}{3k-5}\)
\(\dfrac{2c+d}{3c-5d}=\dfrac{2dk+d}{3dk-5d}=\dfrac{2k+1}{3k-5}\)
Do đó: \(\dfrac{2a+b}{3a-5b}=\dfrac{2c+d}{3c-5d}\)
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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a-5b}{2c-5d}\)
\(\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{2a-5b}{2c-5d}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\left(dpcm\right)\)
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Gọi a/b=c/d=k nên a=bk;c=dk
=>2a+5b/3a-4b=2bk+5b/3bk-4b=b(2k+5)/b(3k-4)=2k+5/3k-4(1)
=>2c+5d/3c-4d=2dk+5d/3dk-4d=d(2k+5)/d(3k-4)=2k+5/3k-4(2)
Từ (1);(2) =>2a+5b/3a-4b=2c+5d/3c-4d
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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a-5b}{2c-5d}.\)
\(\Rightarrow\frac{a}{c}=\frac{2a-5b}{2c-5d}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)