tim a b c biet a/2014=b/3=c/3 va 2015b-2016c-a=-1
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Ta co : \(\frac{a}{2014}=\frac{b}{3}=\frac{c}{2}\Rightarrow\frac{a}{2014}=\frac{2015b}{6045}=\frac{2016c}{4032}\) va 2015b-2016b-a=-1
Ap dung tinh chat day ti so bang nhau :
\(\frac{2015b}{6045}=\frac{2016c}{4032}=\frac{a}{2014}=\frac{2015b-2016b-a}{6045-4032-2014}=-\frac{1}{-1}=1\)
Suy ra : \(\frac{2015b}{6045}=1\Rightarrow b=1.6045:2015=3\)
\(\frac{2016c}{4032}=1\Rightarrow c=1.4032:2016=2\)
\(\frac{a}{2014}=1\Rightarrow a=1.2014=2014\)
**** nhe
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Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
a) Ta có: \(\left(\frac{a+b}{c+d}\right)^3=\left(\frac{bk+b}{dk+d}\right)^3=\left[\frac{b.\left(k+1\right)}{d.\left(k+1\right)}\right]^3=\left(\frac{b}{d}\right)^3\) (1)
\(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\frac{b^3.k^3-b^3}{d^3.k^3-d^3}=\frac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\frac{b^3}{d^3}=\left(\frac{b}{d}\right)^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
b) Ta có:
\(\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\) (1)
\(\frac{2015a^2+2016c^2}{2015b^2+2016d^2}=\frac{2015.\left(bk\right)^2+2016.\left(dk\right)^2}{2015b^2+2016d^2}=\frac{2015.b^2.k^2+2016.d^2.k^2}{2015.b^2+2016.d^2}=\frac{k^2.\left(2015.b^2+2016d^2\right)}{2015b^2+2016d^2}=k^2\left(2\right)\) Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{2015a^2+2016c^2}{2015b^2+2016d^2}\)
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a + b = 3 ; b + c = 5 ; a + c = 4
b + c - a - b = 5 - 3
c - a = 2
Vậy c = (4 + 2) : 2 = 3
a = 4 - 3 = 1
b = 5 - 3 = 2
Vậy a = 1 ; b = 2 ; c = 3
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\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Leftrightarrow\frac{a}{2}=\frac{b}{\frac{3}{2}}=\frac{c}{\frac{4}{3}}=\frac{a-b}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
=> a = 2.30 = 60
b =30. 3/2 = 45
c = 30 . 4/3 =40
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Ta co:a-b=15
=>2(a-b)=30 hay 2a-2b=30
Co:\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)
\(hay\frac{2a}{4}=\frac{2b}{3}=\frac{3c}{4}\)va 2a-2b=30
Ap dung tinh chat cua day ti so bang nhau ta co:
\(\frac{2a}{4}=\frac{2b}{3}=\frac{3c}{4}=\frac{2a-2b}{4-3}=\frac{30}{1}=30\)
Con lai la tu ban nhe
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