số cặp x,y là : 

N :2 = ??

đ/s:.......

số cặp x,y,z là :

N^* :3=?

sai rồi

\(\frac{x}{y}=\frac{10}{9}\text{ }\Rightarrow\text{ }\frac{x}{10}=\frac{y}{9}\text{ }\left(1\right)\)

\(\frac{y}{z}=\frac{3}{4}=\frac{9}{12}\text{ }\Rightarrow\text{ }\frac{y}{9}=\frac{z}{12}\text{ }\left(2\right)\)

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

Do đó : x = 6 . 10 = 60 ; y = 6 . 9 = 54 ' z = 6 . 12 = 72

Ta có: \(\frac{x}{y}=\frac{10}{9}\) và \(\frac{y}{z}=\frac{3}{4}\)

=> \(\frac{y}{z}=\frac{9}{12}\) 

=> \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{87}{13}=6\)

=>\(\frac{x}{10}=6=>x=60\) ; \(\frac{y}{9}=6=>y=54\); \(\frac{z}{12}=6=>z=72\)

vậy x=60 ; y=54 ; z=72

Từ x/y=9/10

suy ra x/10=y/9 (1)

y/z=3/4

suy ra y/3=z/4 (2)

từ (1) và (2) suy ra

x/30=y/27=z/36=x-y+x/30-27+36=78/39=2 (theo tính chất của dãy tỉ số)

suy ra x/30=2 nên x=2*30=60

           y/27=2 nên y=2*27=54

           z/36=2 nên z=2*36=72

          Vậy x=60

                 y=54

                  z=72

chuẩn 100% đó thầy mình dạy hoài

chúc bạn học tốt

Lời giải:

Áp dụng BĐT AM-GM ta có:

\((x-1)+1\geq 2\sqrt{x-1}\Leftrightarrow \frac{x}{2}\geq \sqrt{x-1}\)

\(\Rightarrow yz\sqrt{x-1}\leq \frac{xyz}{2}\)

\((y-4)+4\geq 4\sqrt{y-4}\) \(\Leftrightarrow \frac{y}{4}\geq \sqrt{y-4}\)

\(\Rightarrow zx\sqrt{y-4}\leq \frac{xyz}{4}\)

\((z-9)+9\geq 6\sqrt{z-9}\Leftrightarrow \frac{z}{6}\geq \sqrt{z-9}\)

\(\Rightarrow xy\sqrt{z-9}\leq \frac{xyz}{6}\)

Do đó:

\(Q\leq \frac{\frac{xyz}{2}+\frac{xyz}{4}+\frac{xyz}{6}}{xyz}=\frac{xyz.\frac{11}{12}}{xyz}=\frac{11}{12}\)

Vậy \(Q_{\max}=\frac{11}{12}\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-1=1\\ y-4=4\\ z-9=9\end{matrix}\right.\Leftrightarrow x=2; y=8; z=18\)

a,

\(pt\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)

\(\dfrac{x-2}{2}=\dfrac{y-4}{3}=\dfrac{z-8}{5}\)

\(\Rightarrow\dfrac{x-2}{2}+2=\dfrac{y-4}{3}+2=\dfrac{z-8}{5}+2\)

\(\Rightarrow\dfrac{x+2}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{5}\)

\(\Rightarrow\left(\dfrac{x+2}{2}\right)^2=\left(\dfrac{y+2}{3}\right)^2=\left(\dfrac{z+2}{5}\right)^2\)

\(\Rightarrow\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}=\dfrac{3.\left(y+2\right)^2}{27}\dfrac{\left(x+2\right)^2+3\left(y+2\right)^2-\left(z+2\right)^2}{4+27-25}=\dfrac{24}{6}=4\)\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=16\\\left(y+2\right)^2=36\\\left(z+2\right)^2=100\end{matrix}\right.\)

Bạn chia trường hợp rồi tìm x,y,z nhé

ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

\(\Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{xz+yz+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^8=\left(-y\right)^8\\y^9=\left(-z\right)^9\\z^{10}=\left(-x\right)^{10}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^8-y^8=0\\y^9+z^9=0\\x^{10}-z^{10}=0\end{matrix}\right.\)\(\Rightarrow\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)=0\)

\(\Rightarrow M=\dfrac{3}{4}\)

\(x\left(x+y+z\right)=-5\Rightarrow x+y+z=-\frac{5}{x}\)

\(y\left(x+y+z\right)=9\Rightarrow x+y+z=\frac{9}{y}\)

\(z\left(x+y+z\right)=5\Rightarrow x+y+z=\frac{5}{z}\)

\(\Rightarrow\frac{-5}{x}=\frac{9}{y}=\frac{5}{z}=x+y+z=\frac{-5+9+5}{x+y+z}=\frac{9}{x+y+z}\Rightarrow x+y+z=\frac{9}{x+y+z}\)

\(\Rightarrow\left(x+y+z\right)^2=9\Rightarrow x+y+z=+-3\)

\(x+y+z=3\Rightarrow x=-\frac{5}{3};y=\frac{9}{3}=3;z=\frac{5}{3}\)

\(x+y+z=-3\Rightarrow x=\frac{-5}{-3}=\frac{5}{3};y=\frac{9}{-3}=-3;z=\frac{5}{-3}=-\frac{5}{3}\)

Cộng vế theo vế ta được:

x(x+y+z)+y(x+y+z)+z(x+y+z)=-5+9+5

=>(x+y+z)(x+y+z)=9

=>(x+y+z)²=\(\left(\pm3\right)^2\)

Với x+y+z=3 => \(\hept{\begin{cases}3x=-5\\3y=9\\3z=5\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{5}{3}\\y=3\\z=\frac{5}{3}\end{cases}}}\)

Với x+y+z=-3 => \(\hept{\begin{cases}-3x=-5\\-3y=9\\-3z=5\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-3\\z=-\frac{5}{3}\end{cases}}}\)

Vậy...

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405