\(a+\frac{b}{c}=11\)
\(b+\frac{a}{c}=14\)
\(\frac{a+b}{c}=?\)
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Giải:
Ta có:
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{14}{22}\\\dfrac{c}{d}=\dfrac{11}{13}\\\dfrac{e}{f}=\dfrac{13}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{7}{11}\\\dfrac{c}{d}=\dfrac{11}{13}\\\dfrac{e}{f}=\dfrac{13}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{7}=\dfrac{b}{11}\\\dfrac{c}{11}=\dfrac{d}{13}\\\dfrac{e}{13}=\dfrac{f}{17}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{7}=\dfrac{b}{11}=\dfrac{a+b}{7+11}=\dfrac{M}{18}\left(1\right)\\\dfrac{c}{11}=\dfrac{d}{13}=\dfrac{c+d}{11+13}=\dfrac{M}{24}\left(2\right)\\\dfrac{e}{13}=\dfrac{f}{17}=\dfrac{e+f}{13+17}=\dfrac{M}{30}\left(3\right)\end{matrix}\right.\)
Kết hợp \(\left(1\right);\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow M\in BC\left(18;24;30\right)\)
Mặt khác \(M\) là số tự nhiên nhỏ nhất có 4 chữ số
Nên \(M=1080\)
Vậy \(M=1080\)
Ta có : \(\frac{a}{b}=\frac{14}{22}\Rightarrow\frac{a}{14}=\frac{b}{22}=\frac{a+b}{14+22}=\frac{M}{36}\)
\(\frac{c}{d}=\frac{11}{13}\Rightarrow\frac{c}{11}=\frac{d}{13}=\frac{c+d}{11+13}=\frac{M}{24}\)
\(\frac{e}{f}=\frac{13}{17}\Rightarrow\frac{e}{13}=\frac{f}{17}=\frac{e+f}{13+17}=\frac{M}{30}\)
Nhận thấy M chia hết cho 36,24,30 => \(M⋮36,M⋮24,M⋮30\)
=> \(M\in BC\left(36,24,30\right)\)
Ta có : 36 = 22 . 32
24 = 23 . 3
30 = 2.3.5
=> \(BCNN\left(36,24,30\right)=2^3\cdot3^2\cdot5=360\)
=> \(BC\left(36,24,30\right)=B\left(360\right)=\left\{0;360;720;1080\right\}\)
Vậy số tự nhiên của M là 1080
ta có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)
\(\Rightarrow14.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=14.\frac{1}{7}\)
\(\Rightarrow\frac{14}{a+b}+\frac{14}{b+c}+\frac{14}{c+a}=2\)
mà a+b+c =14
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=2\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2\)
\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2-3\)
\(\Rightarrow A=-1\)
CHÚC BN HỌC TỐT!!!!!!
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
a) Ta có:
\(\frac{11a+3b}{11c+3d}=\frac{11bk+3b}{11dk+3d}=\frac{b\left(11k+3\right)}{d\left(11k+3\right)}=\frac{b}{d}\) (1)
\(\frac{3a-11b}{3c-11d}=\frac{3bk-11b}{3dk-11d}=\frac{b\left(3k-11\right)}{d\left(3k-11\right)}=\frac{b}{d}\) (2)
Từ (1) và (2) suy ra \(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\) (đpcm)
b) Ta có:
\(\frac{1111c-99d}{9999c-11d}=\frac{1111dk-99d}{9999dk-11d}=\frac{d\left(1111k-99\right)}{d\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\) (1)
\(\frac{1111a-99b}{9999a-11b}=\frac{1111bk-99b}{9999bk-11b}=\frac{b\left(1111k-99\right)}{b\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\) (2)
Từ (1) và (2) suy ra \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\) (đpcm)
ĐKXĐ: \(c\ne0\)
Có: \(\hept{\begin{cases}a+\frac{b}{c}=11\\b+\frac{a}{c}=14\end{cases}\Leftrightarrow}a+b+\frac{a+b}{c}=25\)
\(\Leftrightarrow\left(a+b\right)\left(1+\frac{1}{c}\right)=\frac{a+b}{c}\cdot\left(c+1\right)=25\)
Vì \(c+1\ne1\)
nên: \(\frac{a+b}{c}=1\)hoặc \(\frac{a+b}{c}=5\)hoặc \(\frac{a+b}{c}=-5\)