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Ta có: \(3a=7b\Rightarrow\dfrac{a}{7}=\dfrac{b}{3}\)
\(4b=3c\Rightarrow\dfrac{b}{3}=\dfrac{c}{4}\)
Khi đó: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)
Do \(\left\{{}\begin{matrix}\dfrac{a}{7}=30\\\dfrac{4b}{12}=30\\\dfrac{5c}{20}=30\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\).
Từ đề ta có: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)
\(\Rightarrow\left\{{}\begin{matrix}a=30\cdot7\\b=30\cdot3\\c=30\cdot4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)
Vậy..........
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a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}.\)
\(\Rightarrow\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)
b) ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{2c}{2d}=\frac{4a}{4b}\)
Lại có: \(\frac{5a}{5b}=\frac{2c}{2d}=\frac{5a+2c}{5b+2d}\)
\(\Rightarrow\frac{4a}{4b}=\frac{5a+2c}{5b+2d}\Rightarrow\frac{5a+2c}{4a}=\frac{5b+2d}{4b}\)
c) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Lại có: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\frac{\left(a+b^2\right)}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
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Cho 3a = 4b và a - b = 12 thì a
Ta có a- b =12
=> 3a - 3b = 36
=> 3a = 36 + 3b
mad 3a = 4b
=> 4b = 36 + 3b
=> b = 36
a = 48 (do a- b = 12)
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3a = 4b => a/4 = b/3 và a - b = 10
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\frac{a}{4}=\frac{b}{3}=\frac{a-b}{4-3}=\frac{10}{1}=10\)
Suy ra: \(\frac{a}{4}=10\Rightarrow a=10\cdot4=40\)
\(\frac{b}{3}=10\Rightarrow b=10\cdot3=30\)
Vậy a = 40, b = 30
3a=4b=>a=4/3b
=>a-b=4/3b-b=1/3b=10
=>b=10:1/3=30
=>a=30+10=40
vậy (a;b)=(40;30)
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Ta có: \(\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)
\(\Rightarrow\frac{3a+4b}{3a-4b}-1=\frac{3c+4d}{3c-4d}-1\)
\(\Leftrightarrow\frac{8b}{3a-4b}=\frac{8d}{3c-4d}\)
\(\Rightarrow b\left(3c-4d\right)=d\left(3a-4b\right)\)
\(\Leftrightarrow3bc=3ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
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a) \(G=\frac{\frac{3a}{b}-\frac{2b}{b}}{\frac{a}{b}-\frac{3b}{b}}=\frac{3.\frac{10}{3}-2}{\frac{10}{3}-3}=\frac{10-2}{\frac{1}{3}}=24\)
b) \(H_1=\frac{\frac{2a-3b}{b}}{\frac{4a+3b}{b}}=\frac{\frac{2a}{b}-\frac{3b}{b}}{\frac{4a}{b}+\frac{3b}{b}}=\frac{2.\frac{10}{3}-3}{4.\frac{10}{3}+3}=\frac{\frac{11}{3}}{\frac{49}{3}}=\frac{11}{49}\)
\(H_2=\frac{\frac{5a-4b}{b}}{\frac{3a+b}{b}}=\frac{5.\frac{a}{b}-4}{3.\frac{a}{b}+1}=\frac{5.\frac{10}{3}-4}{3.\frac{10}{3}+1}=\frac{\frac{38}{3}}{\frac{33}{3}}=\frac{38}{33}\)
=> \(H=\frac{11}{49}-\frac{38}{33}=\frac{-1499}{1617}\)
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3a = 4b => a/4 = b/3 và a - b = 10
Áp dụng tính chất dãy tỉ số bằng nhau có :
\(\frac{a}{4}\) = \(\frac{b}{3}\) = \(\frac{a-b}{4-3}\) = \(\frac{10}{1}\) = 10
Suy ra :
\(\frac{a}{4}\) = 10 \(\Rightarrow\) a = 10.4 = 40
\(\frac{b}{3}\) = 10 \(\Rightarrow\) b = 10.3 = 30
Vậy a = 40, b = 30
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\(3a=4b=5c=\frac{a}{\frac{1}{3}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{1}{5}}\)
Áp dụng TC DTSBN ta có :
\(\frac{a}{\frac{1}{3}}=\frac{b}{\frac{1}{4}}=\frac{c}{\frac{1}{5}}=\frac{a+b+c}{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}=\frac{94}{\frac{47}{60}}=120\)
=> a = 40 ; b = 30 ; c = 24
a = \(\frac{132}{7}\),
b = \(\frac{99}{7}\)