Tim x:\(\dfrac{x}{y}=\dfrac{1}{3}\)
Biet : \(x-\dfrac{3}{y}=\dfrac{1}{2}\)
Ai lam truoc mk se tick nha!
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CT
2 tháng 8 2017
Mik xin loi, de dung la
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{y}=\dfrac{z}{8}\)va \(3x-2y-z=13\)
1 tháng 6 2022
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
LN
9 tháng 11 2018
a,\(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(x=\dfrac{6}{5}\)
b,\(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\left|x\right|=\dfrac{6}{5}\)
\(\Rightarrow x=\pm\dfrac{6}{5}\)
c,\(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(x=\dfrac{-5.24}{15}\)
\(x=\dfrac{-24}{5}\)
d,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
+\(\dfrac{x}{4}=-21\Rightarrow x=-21.4=-84\)
+\(\dfrac{y}{5}=-21\Rightarrow y=-21.5=-105\)
Vậy x=-84 ; y=-105
9 tháng 11 2018
a/ \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
Vậy...
b/ \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy...
c/ \(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(\Leftrightarrow15x=-120\)
\(\Leftrightarrow x=-8\)
Vậy...
c/ Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-21\\\dfrac{y}{5}=-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-84\\y=-105\end{matrix}\right.\)
Vậy..
24 tháng 2 2018
a) \(\dfrac{x}{2}=-\dfrac{6}{3}=-2\Rightarrow x=2.\left(-2\right)=-4\)
b) \(\dfrac{2}{x}=\dfrac{y}{-3}\Leftrightarrow y=-\dfrac{6}{x}\) y thuộc Z => x thuộc {+-6;+-3;+-2;+-1}
(x;y) =(-6;1);(-3;2); (-2;3);(-1;6) ; (6;-1);(3-2);(2;-3);(1;-6)
KK
20 tháng 3 2017
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2xy\sqrt{x}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2yz\sqrt{y}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2xz\sqrt{z}}=\dfrac{1}{xz}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\) ( 1 )
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2\sqrt{\dfrac{1}{x^2y^2}}=\dfrac{2}{xy}\\\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge2\sqrt{\dfrac{1}{y^2z^2}}=\dfrac{2}{yz}\\\dfrac{1}{z^2}+\dfrac{1}{x^2}\ge2\sqrt{\dfrac{1}{x^2z^2}}=\dfrac{2}{xz}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\)
\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x^3+y^2}+\dfrac{2\sqrt{y}}{y^3+z^2}+\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\) ( đpcm )
18 tháng 1 2016
mk ko co thoi gian dua dau
ai lam ca loi giai mk pick cho
BC
1
DH
13 tháng 10 2017
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=>4x=6y=>x=\dfrac{3y}{2}\)\(=>4z=3y=>z=\dfrac{3y}{4}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3< =>\dfrac{1}{\dfrac{3y}{2}}+\dfrac{1}{y}+\dfrac{1}{\dfrac{3y}{4}}=3\)
\(< =>\dfrac{2}{3y}+\dfrac{1}{y}+\dfrac{4}{3y}=3< =>\dfrac{2}{3y}+\dfrac{3}{3y}+\dfrac{4}{3y}=3\)
\(< =>\dfrac{9}{3y}=3< =>\dfrac{3}{y}=3< =>3=3y=>y=1\)
\(=>x=\dfrac{3y}{2}=\dfrac{3}{2};z=\dfrac{3y}{4}=\dfrac{3}{4}\)
ta có : x:\(\dfrac{x}{y}\)=\(\dfrac{1}{3}\)
->x.\(\dfrac{y}{x}\)=\(\dfrac{1}{3}\)
->y=\(\dfrac{1}{3}\)
->x-\(\dfrac{3}{\dfrac{1}{3}}\)=\(\dfrac{1}{2}\)
->x = \(\dfrac{19}{2}\)
Vậy......
mình có mà, mình thay luôn vào, bạn nhìn ở dấu -> thứ ba ý