x:y:z=4:5:6

--> x/4=y/5=z/6

Đặt x=4k; y=5k; z=6k

x^2-2y^2+z^2=18

(4k)^2-2.(5k)^2+(6k)^2=18

2k^2=18

k^2=9

k=3 hoặc k=-3

Khi k=3

--> x=4.3=12

y=5.3=15

z=6.3=18

Khi k=-3

--> x=4.(-3)=-12

y=5.(-3)=-15

z=6.(-3)=-18

Ta có: 4x = 7y

<=> \(\dfrac{4}{y}=\dfrac{7}{x}\)

<=> \(\dfrac{16}{y^2}=\dfrac{49}{x^2}\)

=> \(\dfrac{16+49}{x^2+y^2}=\dfrac{65}{260}=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}x=28\\y=16\end{matrix}\right.\)

Lời giải:

$4x=7y\Rightarrow \frac{x}{7}=\frac{y}{4}=k$

$\Rightarrow x=7k; y=4k$

Ta có: $x^2+y^2=260$

$\Leftrightarrow (7k)^2+(4k)^2=260$

$\Leftrightarrow  65k^2=260$

$\Leftrightarrow k^2=4$

$\Rightarrow k=\pm 2$
Nếu $k=2$ thì $x=7k=14; y=4k=8$. Ta có $(x,y)=(14,8)$

Nếu $k=-2$ thì $x=7k=-14; y=4k=-8$. Ta có $(x,y)=(-14,-8)$

Đặt x=2k; y=4k

x^2.y^2=4 

(2k)^2.(4k)^2=4

64.k^4=4

k^4=1/16

k=1/2 hoặc k=-1/2

Khi k=1/2 --> x=2.1/2=1

y=4.1/2=2

Khi k=-1/2 --> x=2.(-1/2)=-1

y=4.(-1/2)=-2

c) \(2x=3y=5z\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒\(\left\{{}\begin{matrix}x=5.15=75\\y=5.10=50\\z=5.6=30\end{matrix}\right.\)

Lời giải:

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$\Rightarrow \frac{x}{21}=\frac{y}{14}$

$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}$

Vậy:

$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$

$\Rightarrow x=21.2=42; y=14.2=28; z=10.2=20$

a) Áp dụng t/c dtsbn:

 \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

b) \(\dfrac{3}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)

Và \(x+16=y\Rightarrow y-x=16\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{y-x}{7-3}=\dfrac{16}{4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)

a) \(\dfrac{x}{7}=\dfrac{y}{13}\)

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

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

⇒\(\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

~ Chúc bạn ho tốt ~ !!!!!!!!!!!!

nhanh vậy 

ai làm cho mik đi

\(a)\)Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2\cdot(2x+3)-(4x+5)}{2\cdot(5x+2)-(10x+2)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra :

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow1\cdot(5x+2)=2\cdot(2x+3)\)

\(5x+2=4x+6\)

\(5x-4x=6-2\)

\(x=4\)

\(b)\)Ta có : \(\frac{4}{x-3}=\frac{8}{y-6}=\frac{20}{z-15}\)

\(\Rightarrow\frac{x-3}{4}=\frac{y-6}{8}=\frac{z-15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{6}{8}=\frac{z}{20}-\frac{15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{3}{4}=\frac{z}{20}-\frac{3}{4}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{8}=\frac{z}{20}\)

Đặt : \(\frac{x}{4}=\frac{y}{8}=\frac{z}{20}=k\Rightarrow x=4k;y=8k;z=20k\)

Thay vào đề , ta có : xyz = 640

\(\Rightarrow4k\cdot8k\cdot20k=640\)

\(\Rightarrow640k^3=640\)

\(\Rightarrow k^3=1\)

\(\Rightarrow k=1\)

\(\Rightarrow x=4;y=8;z=20\)

Vậy

cam on nha