ta co 6/11.x=20

x=20.11/6

x=110/3

ta co 9/2y=20 

y=20.2/9=40/9

ta co 18/5 z=20

z= 20.5/18

z=50/9

\(x=20:\frac{6}{11}=\frac{110}{3}\)

\(y=20:\frac{9}{2}=\frac{40}{9}\)

\(z=20:\frac{18}{5}=\frac{50}{9}\)

a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)

8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)

=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)

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

\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)

=> x = 24,y = 15,z = 6

b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)

\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)

=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

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

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)

=> x = -165 , y = -20 , z = -25

c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k

=> xyz = 12k . 9k . 5k

=> xyz = 540k³

=> 540k³ =20

=> k³ = 20/540

=> k³ = 1/27

=> k = 1/3

Do đó : x= 4 , y = 3 , z = 5/3

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) va -x+y+z=-120

\(\left(\frac{6}{11}\right)x=\left(\frac{9}{2}\right)y\Rightarrow x=\left(\frac{9}{2}\right)\left(\frac{11}{6}\right)y=\left(\frac{33}{4}\right)y\)

\(\left(\frac{9}{2}\right)y=\left(\frac{18}{5}\right)z\Rightarrow z=\left(\frac{9}{2}\right)\left(\frac{5}{18}\right)y=\left(\frac{5}{4}\right)y\)

\(\Rightarrow-x+y+z=\left(\frac{-33}{4}\right)y+y+\left(\frac{5}{4}\right)y=-120\)

\(\Rightarrow-6y=-120\Rightarrow y=20;x=165;z=25\)

x + y + z = - 90

bn ghi vào fx cho dễ nhìn chứ nhìn như này khó quá

sợ lm sai đề ^^

Có phải đề như thế này không Hoàng Ngọc Phương ?

\(\dfrac{6}{11}\) . \(x\) \(=\) \(\dfrac{9}{2}\) . \(y\) \(=\) \(\dfrac{18}{5}\) . \(z\) và \(x\) \(+\) \(y\) \(+\) \(z\) \(=\) \(-120\)

\(\dfrac{6}{11}.x=\dfrac{9}{2}.y=\dfrac{18}{5}.z\)

\(\Rightarrow\dfrac{6}{11}.x.\dfrac{1}{18}=\dfrac{9}{2}.y.\dfrac{1}{18}=\dfrac{18}{5}.z.\dfrac{1}{18}\)

\(\Rightarrow\dfrac{x}{33}=\dfrac{y}{4}=\dfrac{z}{5}\)

\(\Rightarrow\dfrac{-x}{-33}=\dfrac{y}{4}=\dfrac{z}{5}\)

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

\(\dfrac{-x}{-33}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{-x+y+z}{-33+4+5}=\dfrac{-120}{-24}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.33=165\\y=5.4=20\\z=5.5=25\end{matrix}\right.\)

Ta có : \(\left\{{}\begin{matrix}x-y=-9\\y-z=-10\\z+x=11\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\z+x=11\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\y-9+10+y=11\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\2y+1=11\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=5-9=-4\\z=10+5=15\\y=5\end{matrix}\right.\)

Vậy các giá trị x; y; z lần lượt là -4; 5; 15 .

Ta có:

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

\(\Rightarrow x-z=-19\) ; \(z+x=11\)

\(\Rightarrow\left(x-z\right)+\left(z+x\right)=-19+11\)

\(\Rightarrow2x=-8\Rightarrow x=-4\)

\(\Rightarrow y=5;z=15\)