a, Ta có : 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

mà 2x+3y-z=50

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

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

`P=x^3/(x+y)+y^3/(y+z)+z^3/(z+x)`

`=x^4/(x^2+xy)+y^4/(y^2+yz)+z^4/(z^2+zx)`

Ad bđt cosi-swart:

`P>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+xy+yz+zx)`

Mà `xy+yz+zx<=x^2+y^2+z^2)`

`=>P>=(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2))=(x^2+y^2+z^2)/2=3/2`

Dấu "=" xảy ra khi `x=y=z=1`

`Q=(x^3+y^3)/(x+2y)+(y^3+z^3)/(y+2z)+(z^3+x^3)/(z+2x)`

`Q=(x^3/(x+2y)+y^3/(y+2z)+z^3/(z+2x))+(y^3/(x+2y)+z^3/(y+2z)+x^3/(z+2x))`

`Q=(x^4/(x^2+2xy)+y^4/(y^2+2yz)+z^4/(z^2+2zx))+(y^4/(xy+2y^2)+z^4/(yz+2z^4)+x^4/(xz+2x^2))`

Áp dụng BĐT cosi-swart ta có:

`Q>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+2xy+2yz+2zx)+(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2)+xy+yz+zx))`

Mà`xy+yz+zx<=x^2+y^2+z^2`

`=>Q>=(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))+(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2)^2)/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2))/3=2`

Dấu "=" xảy ra khi `x=y=z=1.`

Ta có: \(\dfrac{x}{-5}=\dfrac{y}{3}\)

nên \(\dfrac{x}{-10}=\dfrac{y}{6}\left(1\right)\)

Ta có: \(\dfrac{x}{2}=\dfrac{z}{3}\)

nên \(\dfrac{x}{-10}=\dfrac{z}{-15}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{-10}=\dfrac{y}{6}=\dfrac{z}{-15}\)

hay \(\dfrac{2x}{-20}=\dfrac{3y}{18}=\dfrac{z}{-15}\)

mà 2x+3y=1

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{-20}=\dfrac{3y}{18}=\dfrac{z}{-15}=\dfrac{2x+3y}{-20+18}=\dfrac{1}{-2}=-\dfrac{1}{2}\)

Do đó: \(\left\{{}\begin{matrix}2x=10\\3y=-6\\z=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-2\\z=30\end{matrix}\right.\)

Mn ơi giúp mk với , please !!!

1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)

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

\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)

=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)

2. Ta có:

- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

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

\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\Rightarrow\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}=\dfrac{2x+4-3y+15+z+1}{6-\left(-12\right)+5}=\dfrac{\left(2x-3y+z\right)+\left(4+15+1\right)}{23}=\dfrac{72+20}{23}=\dfrac{92}{23}=4\)

\(\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\\ \dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\\ \dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

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

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2x-3y+z+4+15+1}{2\cdot3-3\cdot\left(-4\right)+5}=\dfrac{92}{23}=4\)

Do đó: x=10; y=-11; z=4

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\text{ và }2x-3y+z=72\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2\left(x+2\right)-3\left(y-5\right)+z+1}{2.3-3.\left(-4\right)+5}=\dfrac{92}{23}=4\)

\(\Rightarrow\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\)

\(\dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\)

\(\dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

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