Đề sai sao á :))

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

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

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. 

$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$

$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$

$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$

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.

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)

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 )

bạn nhấn vào  đúng 0 sẽ ra đáp án

mk ko co thoi gian dua dau

ai lam ca loi giai mk pick cho

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=>4x=6y=>x=\dfrac{3y}{2}\)\(=>4z=3y=>z=\dfrac{3y}{4}\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3< =>\dfrac{1}{\dfrac{3y}{2}}+\dfrac{1}{y}+\dfrac{1}{\dfrac{3y}{4}}=3\)

\(< =>\dfrac{2}{3y}+\dfrac{1}{y}+\dfrac{4}{3y}=3< =>\dfrac{2}{3y}+\dfrac{3}{3y}+\dfrac{4}{3y}=3\)

\(< =>\dfrac{9}{3y}=3< =>\dfrac{3}{y}=3< =>3=3y=>y=1\)

\(=>x=\dfrac{3y}{2}=\dfrac{3}{2};z=\dfrac{3y}{4}=\dfrac{3}{4}\)