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Theo đề bài, ta có:
\(\dfrac{3x}{4}=\dfrac{y}{2}=\dfrac{3z}{5}\) và x - z = 15
\(\Rightarrow\dfrac{3x}{4}=\dfrac{y}{2}\Rightarrow6x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}\) (1)
\(\Rightarrow\dfrac{y}{2}=\dfrac{3z}{5}\Rightarrow5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\) (2)
(1)(2) \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}=\dfrac{x-z}{4-5}=-\dfrac{15}{1}=-15\)
\(\Rightarrow x=-60;y=-90;z=-75\)
\(\Rightarrow x+y+z=-225\)
Chả hỉu gì cả.
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
Nếu:
\(\dfrac{a+b}{a}=\dfrac{c+d}{c}\Leftrightarrow c\left(a+b\right)=a\left(c+d\right)\)
\(ac+bc=ac+ad\)
\(bc=ad\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\rightarrowđpcm\)
Đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k
=> a=k.b ; c=k.d
Ta có :
\(\dfrac{a+b}{a}\)=\(\dfrac{b.k+b}{b}\)=\(\dfrac{b.\left(k+1\right)}{b}\)=k+1 ( 1 )
\(\dfrac{c+d}{c}\)=\(\dfrac{d.k+d}{d}\)=\(\dfrac{d.\left(k+1\right)}{d}\)=k+1 ( 2 )
Từ (1) và (2) thì : \(\dfrac{a+b}{a}\)=\(\dfrac{c+d}{c}\)
Ta có:\(\frac{x+y}{2}=\frac{y-5}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:\(\frac{x+y}{2}=\frac{y-5}{3}=\frac{x+y+y-5}{2+3}=\frac{x+2y-5}{5}\)
\(\Rightarrow\frac{x+2y-5}{5}=\frac{x+2y-5}{y-1}\)\(\Rightarrow y-1=5\Rightarrow y=6\)
\(\Rightarrow\frac{x+6}{2}=\frac{6-5}{3}\)\(\Rightarrow\frac{x+6}{2}=\frac{1}{3}\)
\(\Rightarrow3\cdot\left(x+6\right)=2\)
\(\Rightarrow3x+18=2\)
\(\Rightarrow3x=-16\Rightarrow x=\frac{-16}{3}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x+y}{2}=\frac{y-5}{3}=\frac{x+y+y-5}{2+3}=\frac{x+2y-5}{5}\)
\(=\frac{x+2y-5}{y-1}\) (theo đề bài)
=> y - 1 = 5
=> y = 5 + 1 = 6
Thay y = 6 vào đề bài ta có: \(\frac{x+6}{2}=\frac{7-6}{3}=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}.2-6=\frac{-16}{3}\)
Vậy \(x=\frac{-16}{3};y=6\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
Vậy nếu \(\dfrac{a}{b}=\dfrac{b}{c}\) thì \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\left(đpcm\right)\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{x-y-1+3}{4011}=\dfrac{4009-1+3}{4011}=\dfrac{4011}{4011}=1.\)
Từ đó:
\(\dfrac{x-1}{2005}=1\Rightarrow x-1=2005\Rightarrow x=2006.\)
\(\dfrac{3-y}{2006}=1\Rightarrow3-y=2006\Rightarrow y=-2003.\)
Vậy \(x=2006;y=-2003.\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\\ \dfrac{2k}{2}=\dfrac{3k}{3}=\dfrac{4k}{4}\\ \Rightarrow\dfrac{\left(2k\right)^2}{2^2}=\dfrac{\left(3k\right)^2}{3^2}=\dfrac{2\left(4k\right)^2}{2\cdot4^2}\\ \Leftrightarrow\dfrac{4k^2}{4}=\dfrac{9k^2}{9}=\dfrac{32k^2}{32}=\dfrac{4k^2-9k^2+32k^2}{4-9+32}=\dfrac{108}{27}=4\\ \dfrac{4k^2-9k^2+32k^2}{4-9+32}=4\\ \Rightarrow\dfrac{\left(4-9+32\right)k^2}{4-9+32}=4\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ k=2\Rightarrow\left\{{}\begin{matrix}a=2k=2\cdot2=4\\b=3k=3\cdot2=6\\c=4k=4\cdot2=8\end{matrix}\right.\\ k=-2\Rightarrow\left\{{}\begin{matrix}a=2k=2\cdot\left(-2\right)=-4\\b=3k=3\cdot\left(-2\right)=-6\\c=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)
Vậy ...
Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau có :
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{a}{2}=4\\\dfrac{b}{3}=4\\\dfrac{c}{4}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=8\\b=12\\c=16\end{matrix}\right.\)
\(\left(\dfrac{-5}{13}\right)^{2017}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(-\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(-\dfrac{5}{13}\right)\cdot\left[\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}\right]=\left(-\dfrac{5}{13}\right)\cdot1^{2016}=\left(-\dfrac{5}{13}\right)\cdot1=-\dfrac{5}{13}\)
Vì \(\dfrac{x}{2}=\dfrac{y}{5}\Rightarrow x=2k;y=5k\) (1)
Thay \(x\cdot y=10\) vào (1), ta có:
\(2k\cdot5k=10\)
\(\Rightarrow10k^2=10\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
Nếu \(k=1\) thì: \(\left\{{}\begin{matrix}x=2\cdot1=2\\y=5\cdot1=5\end{matrix}\right.\)
Nếu \(k=-1\) thì \(\left\{{}\begin{matrix}x=2\cdot\left(-1\right)=-2\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)
Đặt :
\(\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
Thay \(\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\) vào \(x.y=10\) ta được :
\(x.y=2k.5k=10\)
\(\Leftrightarrow10k^2=10\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}k^2=1^2\\k^2=\left(-1\right)^2\end{matrix}\right.\)
+) \(k=1\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
+) \(k=-1\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-5\end{matrix}\right.\)
Vậy ..