a,Áp sụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\)  (Vì 3x-2z=15)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)

Vậy ...
 

b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\)    \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)

Vậy ...

c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\)                                                         (Vì 3z-2x=36)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\)     \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)

Vậy ... 

d: x/2=y/1=z/3

mà 3x+4z=16+2=18

nên Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3}=\dfrac{3x+4z}{3\cdot2+4\cdot3}=\dfrac{18}{18}=1\)

=>x=2; y=1; z=3

a: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{3\cdot3-2\cdot7}=\dfrac{15}{-5}=-3\)

=>x=-9; y=-15; z=-21

b: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x-3y}{2\cdot5-3\cdot3}=\dfrac{100}{1}=100\)

=>x=500; y=300; z=200

c: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{4}=\dfrac{3z-2x}{3\cdot4-2\cdot3}=\dfrac{36}{6}=6\)

=>x=18; y=30; z=24

1.Tìm x, y.

2.TÌM x, y, z.

3.TÌM x, y, z.

1) \(\dfrac{x}{3}=\dfrac{y}{4}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

\(\Rightarrow xy=12k^2=192\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm16\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\end{matrix}\right.\)

2) Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{-90}{9}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).2=-20\\y=\left(-10\right).3=-30\\z=\left(-10\right).5=-50\end{matrix}\right.\)

3) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.8=16\\z=2.5=10\end{matrix}\right.\)

Theo đề : \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2+y^2+2z^2=108\)

\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\left(\dfrac{z}{4}\right)^2\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=2.\left(\dfrac{z}{4}\right)^2=>\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2+y^2+2z^2}{4+9+32}=\dfrac{108}{45}=\dfrac{12}{5}\)

Với \(\dfrac{x^2}{2}=\dfrac{12}{5}\Rightarrow x^2=\dfrac{12}{5}.2=\dfrac{24}{5}\Rightarrow x=\dfrac{2\sqrt{30}}{5}\)

\(\dfrac{y^2}{3}=\dfrac{12}{5}\Rightarrow y^2=\dfrac{12}{5}.3=\dfrac{36}{5}\Rightarrow y=\dfrac{6\sqrt{5}}{5}\)

\(\dfrac{2z^2}{4}=\dfrac{12}{5}\Rightarrow2z^2=\dfrac{12}{5}.4=\dfrac{48}{5}\Rightarrow z^2=\dfrac{24}{5}=>\dfrac{2\sqrt{30}}{5}\)

Ta có: \(\dfrac{x}{2}=\dfrac{x^2}{4}\) ; \(\dfrac{y}{3}=\dfrac{y^2}{9}\) ; \(\dfrac{z}{4}=\dfrac{2z^4}{32}\)

Ta có: \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)

\(x=4.2=8\)
\(y=4.3=12\)
\(z=4.32=128\)

Vậy 3 số cần tìm là:
x = 8; y = 12; z=128

x=(+,-)4,y=(+,-)6,z=(+,-)8

mị chỉ làm ngắn gọn thoi!!!!!!!

Theo đề ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2-y^2+2z^2=108\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2}{2^2}=\dfrac{y^2}{3^2}=\dfrac{2z^2}{2.4^2}=\dfrac{x^2-y^2+2x^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\dfrac{x}{2}=4\Rightarrow x=4.2=8\)

\(\dfrac{y}{3}=4\Rightarrow y=4.3=12\)

\(\dfrac{z}{4}=4\Rightarrow z=4.4=16\)

Vậy \(x=8\\ y=12\\ z=16\)

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36

Nên theo tính chất của dãy tỉ số bằng nhau ta có:

 \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6

 \(\Rightarrow\)x=6.5=30

     y=6.6=36

     z=6.7=42

vậy x=30,y=36,z=42

`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.`

đúng rùi đó

Sai

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{-21}\)

Áp dugj tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{-21}=\dfrac{x+y+z}{10+15+\left(-21\right)}=\dfrac{92}{14}=\dfrac{46}{7}\)

Còn lại bạn tự tính nha

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\)

⇒\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{2z}{-4}\)

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

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{2z}{-4}=\dfrac{x+y-2z}{5+1-\left(-4\right)}=\dfrac{160}{10}=16\)

⇒\(\left\{{}\begin{matrix}x=16.5=80\\y=16.1=16\\z=16.-2=-32\end{matrix}\right.\)