Tim x:\(\dfrac{x}{y}=\dfrac{1}{3}\)
Biet : \(x-\dfrac{3}{y}=\dfrac{1}{2}\)
Ai lam truoc mk se tick nha!
Tim x,y va z biet
\(\dfrac{x}{3}=\dfrac{y}{y};\dfrac{y}{6}=\dfrac{z}{8}\)va \(3x-2y-z=13\)
Ai lam dung mik tick cho
Bạn ơi đề có sai ko
Sao lại \(\dfrac{y}{y}\)
- Theo đề bài ta có:
\(\dfrac{x}{3}=\dfrac{y}{4},\dfrac{y}{6}=\dfrac{z}{8}\)
=> \(\dfrac{x}{18}=\dfrac{y}{24}=\dfrac{z}{32}\)
=> \(\dfrac{3x}{54}=\dfrac{2y}{48}=\dfrac{z}{32}\)
- Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{54}=\dfrac{2y}{48}=\dfrac{z}{32}\)=\(\dfrac{3x-2y-z}{54-48-32}\)=\(\dfrac{13}{-26}=-2\)
- Suy ra:
x = \(\dfrac{-2.54}{3}=-36\)
y = \(\dfrac{-2.48}{2}=-48\)
z = \(-2.32=-64\)
- Vậy x = -36; y = -48; z = -64
Tim x, biet:
a, \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
b, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
c, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}\)\(=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
d, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Help me!!!!!!!!!!
Can gap lam. Ai lam duoc cau no thi lam nha. Cam on nhieu truoc!!!!!!!!!!!!
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
Giari ptr
a/2x(3x-1)=6x^2-13
b/\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)
Giups mk vs ạ ai nhanh mk tick nha ><
a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)
b: =>2x-2x-1=x-6x
=>-5x=-1
hay x=1/5
Lời giải:
a.
$2x(3x-1)=6x^2-13$
$\Leftrightarrow 6x^2-2x=6x^2-13$
$\Leftrightarrow 2x=13$
$\Leftrightarrow x=\frac{13}{2}$
b.
$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$
$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$
$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$
Tim x, y biet:
a) \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
b) \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
c) \(\dfrac{x}{-5}=\dfrac{24}{15}\)
d) \(\dfrac{X}{4}=\dfrac{Y}{5}\)va \(x-y=21\)
GIUP MK T ^ T
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
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..
a, \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(x\) \(=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
b, \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\left|x\right|\) \(=\dfrac{2}{5}+\dfrac{4}{5}=\dfrac{6}{5}\)
=> \(x=\pm\dfrac{6}{5}\)
c, \(\dfrac{x}{-5}=\dfrac{24}{15}\)
=> \(x:\left(-5\right)=24:15\)
=> \(x\) \(=\left(24:15\right).\left(-5\right)=-8\)
d, \(\dfrac{x}{4}=\dfrac{y}{5}\) và \(x-y=21\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
Suy ra:
\(\dfrac{x}{4}=-21\) => \(\left(-21\right).4=-84\)
\(\dfrac{y}{5}=-21\) => \(\left(-21\right).5=-105\)
Vậy x = -84 và y = -105.
1. tim cac so nguyen x, y biet
a.\(\dfrac{x}{2}\)bang\(\dfrac{-6}{3}\)
b. \(\dfrac{2}{x}\)bang \(\dfrac{y}{-3}\)
c. \(\dfrac{x}{y}\)bang\(\dfrac{-3}{11}\)
d. \(\dfrac{x}{y-1}\)bang\(\dfrac{5}{-19}\)
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)
tim cap so x,y biet:
\(\dfrac{1+3.y}{12}\)\(=\dfrac{1+5.y}{5.x}\)\(=\dfrac{1+7.y}{4.x}\)
Cho x , y , z > 0
Chứng minh rằng \(\dfrac{2\sqrt{x}}{x^3+y^2}+\dfrac{2\sqrt{y}}{y^3+z^2}+\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)
Ai đó giúp tui nhanh nha , thanks you
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 )
tim cac so nguyen x,y biet :
a) (x-1).(y+2)=7
b) x(y-3)=-12
c) (x-3)(y-3)=9
cac ban lam het ra cho minh nha,mk se tick cho!
mk ko co thoi gian dua dau
ai lam ca loi giai mk pick cho
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}biet\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=>4x=6y=>x=\dfrac{3y}{2}\)\(=>4z=3y=>z=\dfrac{3y}{4}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3< =>\dfrac{1}{\dfrac{3y}{2}}+\dfrac{1}{y}+\dfrac{1}{\dfrac{3y}{4}}=3\)
\(< =>\dfrac{2}{3y}+\dfrac{1}{y}+\dfrac{4}{3y}=3< =>\dfrac{2}{3y}+\dfrac{3}{3y}+\dfrac{4}{3y}=3\)
\(< =>\dfrac{9}{3y}=3< =>\dfrac{3}{y}=3< =>3=3y=>y=1\)
\(=>x=\dfrac{3y}{2}=\dfrac{3}{2};z=\dfrac{3y}{4}=\dfrac{3}{4}\)