Ta có x²-xy+y²=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)

=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...

Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :

S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\)  (Netbit)

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)

Bài này dễ thôi:vv

Theo đề ta có: \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\Leftrightarrow\dfrac{xbc+yac+zab}{abc}=0\Leftrightarrow xbc+yac+zab=0\)

Lại có:\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ca}{xz}\right)=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+cay}{xyz}\right)=4\)

=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2.0=4\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=2\)

Vậy...

Lời giải:
\(A=\left(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}\right)\left(\frac{1}{y-z}+\frac{1}{z-x}+\frac{1}{x-y}\right)-\frac{x}{(y-z)(z-x)}-\frac{x}{(y-z)(x-y)}-\frac{y}{(z-x)(x-y)}-\frac{y}{(z-x)(y-z)}-\frac{z}{(x-y)(y-z)}-\frac{z}{(x-y)(z-x)}\)

\(=0-\frac{x(x-y)+x(z-x)+y(y-z)+y(x-y)+z(z-x)+z(y-z)}{(x-y)(y-z)(z-x)}\)

\(=0-\frac{x^2+xz+y^2+xy+z^2+zy-(xy+x^2+yz+y^2+zx+z^2)}{(x-y)(y-z)(z-x)}=0-\frac{0}{(x-y)(y-z)(z-x)}=0\)

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Do đó: x=-70; y=-135; z=-84

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

a) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{5}=\dfrac{z}{4}\end{matrix}\right.\)

 \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+z-y}{10+12-15}=-\dfrac{49}{7}=-7\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{-2}\\\dfrac{x}{6}=\dfrac{z}{7}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{-4}=\dfrac{z}{7}=\dfrac{3x}{18}=\dfrac{2y}{-8}=\dfrac{3x-z+2y}{18-7-8}=\dfrac{3}{3}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.6=6\\y=1.\left(-4\right)=-4\\z=1.7=7\end{matrix}\right.\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)