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23 tháng 7 2017
Ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+c}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{3a^6+c^6}{3b^6+d^6}\)(2)
Từ (1) ; (2) \(\Rightarrow\frac{\left(a+c\right)^6}{\left(b+d\right)^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (đpcm)
MV
19 tháng 7 2017
a,
\(a+b=-9\\ b+c=2\\ c+a=-3\\ \Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\\ 2a+2b+2c=-10\\ 2\left(a+b+c\right)=-10\\ a+b+c=-5\\ a+b=-9\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-9\right)+c=-5\Rightarrow c=4\\ b+c=2\\ \Rightarrow a+b+c=-5\Leftrightarrow a+2=-5\Rightarrow a=-7\\ c+a=-3\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-3\right)+b=-5\Rightarrow b=-2\)
Vậy \(a=-7;b=-2;c=5\)
b,
\(a+b=\dfrac{1}{2}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{-5}{6}\\ \Rightarrow a+b+b+c+c+a=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{-5}{6}\\ 2a+2b+2c=\dfrac{6}{12}+\dfrac{9}{12}+\dfrac{-10}{12}\\ 2\left(a+b+c\right)=\dfrac{5}{12}\\ a+b+c=\dfrac{5}{24}\\ a+b=\dfrac{1}{2}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{2}+c=\dfrac{5}{24}\Rightarrow c=\dfrac{-7}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{5}{24}\Rightarrow a=\dfrac{-13}{24}\\ a+c=\dfrac{-5}{6}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow b+\dfrac{-5}{6}=\dfrac{5}{24}\Rightarrow b=\dfrac{25}{24}\)
Vậy \(a=\dfrac{-13}{24};b=\dfrac{25}{24};c=\dfrac{-7}{24}\)
c,
\(a+b=2\\ b+c=6\\ c+a=3\\ \Rightarrow a+b+b+c+c+a=2+6+3\\ 2a+2b+2c=11\\ 2\left(a+b+c\right)=11\\ a+b+c=5,5\\ a+b=2\\ \Rightarrow a+b+c=5,5\Leftrightarrow2+c=5,5\Rightarrow c=3,5\\ b+c=6\\ \Rightarrow a+b+c=5,5\Leftrightarrow a+6=5,5\Rightarrow a=-0,5\\ c+a=3\\ \Rightarrow a+b+c=5,5\Leftrightarrow b+3=5,5\Rightarrow b=2,5\)
Vậy \(a=-0,5;b=2,5;c=3,5\)
d,
\(a+b=\dfrac{5}{6}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+b+c+c+a=\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{5}{3}\\ 2a+2b+2c=\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{20}{12}\\ 2\left(a+b+c\right)=\dfrac{13}{4}\\ a+b+c=\dfrac{13}{8}\\ a+b=\dfrac{5}{6}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow\dfrac{5}{6}+c=\dfrac{13}{8}\Rightarrow c=\dfrac{19}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{13}{8}\Rightarrow a=\dfrac{7}{8}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow b+\dfrac{5}{3}=\dfrac{13}{8}\Rightarrow b=\dfrac{-1}{24}\)
Vậy \(a=\dfrac{7}{8};b=\dfrac{-1}{24};c=\dfrac{19}{24}\)
19 tháng 7 2017
\(\left\{{}\begin{matrix}a+b=-9\\b+c=2\\c+a=-3\end{matrix}\right.\)
\(\Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\)
\(\Rightarrow2a+2b+2c=-10\)
\(\Rightarrow2\left(a+b+c\right)=-10\)
\(\Rightarrow a+b+c=-5\)
\(\Rightarrow\left\{{}\begin{matrix}c=-5-9=-14\\a=-5-2=-7\\b=-5-\left(-3\right)=-2\end{matrix}\right.\)
S
5 tháng 8 2019
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow a=b=c\Rightarrow M=1\)
17 tháng 7 2017
\(\dfrac{4}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{x}-\dfrac{2y}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{1+2y}{6}\)
\(\Rightarrow24=x\left(1+2y\right)\)
\(\Rightarrow x;1+2y\inƯ\left(24\right)\)
\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)
Mà 1+2y lẻ nên:
\(\left\{{}\begin{matrix}1+2y=1\Rightarrow2y=0\Rightarrow y=0\\x=24\\1+2y=-1\Rightarrow2y=-2\Rightarrow y=-1\\x=-24\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1+2y=3\Rightarrow2y=2\Rightarrow y=1\\x=8\\1+2y=-3\Rightarrow2y=-4\Rightarrow y=-2\\x=-8\end{matrix}\right.\)
5 tháng 11 2019
Bài 2/a
Giả sử \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\Rightarrow\hept{\begin{cases}a=2k\\b=3k\\c=5k\end{cases}}\)
\(\Rightarrow\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)
\(\Rightarrow\frac{3\cdot2k-2\cdot3k}{5}=\frac{2\cdot5k-5\cdot2k}{3}=\frac{5\cdot3k-3\cdot5k}{2}\)
\(\Rightarrow\frac{6k-6k}{5}=\frac{10k-10k}{3}=\frac{15k-15k}{2}\)
\(\Rightarrow\frac{0}{5}=\frac{0}{3}=\frac{0}{2}=0\left(đpcm\right)\)
Bài 2/c
Có a = 2k ; b = 3k ; c = 5k
=> 2 (a - b) (b - c) = a2
=> 2 (2k - 3k) (3k - 5k) = (2k)2
=> 2 (-1)k . (-2)k = 4k2
=> 4k2 = 4k2 (đpcm)
Mình chỉ làm được có vậy thôi, mong bạn thông cảm =))
Chúc bạn học tốt =))
YN
3 tháng 12 2019
\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)
\(\Rightarrow\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{15a-10b}{25}=0\\\frac{6c-15a}{9}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3a-2b=0\\2c-5a=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3a=2b\\2c=5a\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=\frac{b}{3}\\\frac{c}{5}=\frac{a}{2}\end{cases}}\)
\(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)
C
0
3 tháng 8 2017
Từ \(a+b+c=6\Rightarrow\hept{\begin{cases}a+b=6-c\\b+c=6-a\\a+c=6-b\end{cases}}\)
\(\Rightarrow A=\frac{b+c+5}{a+1}+\frac{c+a+4}{b+2}+\frac{a+b+3}{c+3}\)
\(=\frac{6-a+5}{a+1}+\frac{6-b+4}{b+2}+\frac{6-c+3}{c+3}\)
\(=\frac{11-a}{a+1}+\frac{10-b}{b+2}+\frac{9-c}{c+3}\)
\(=-1+\frac{12}{a+1}-1+\frac{12}{b+2}-1+\frac{12}{c+3}\)
\(=-3+12\left(\frac{1}{a+1}+\frac{1}{b+2}+\frac{1}{c+3}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwrarz dưới dạng Engel ta có :
\(A\ge-3+12.\frac{\left(1+1+1\right)^2}{6+\left(a+b+c\right)}=-3+12.\frac{9}{12}=6\) (đpcm)