\(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)

x=-3.17=-51

y=-3.3=-9

câu tiếp nha:\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

x=19.2=38

y=21.2=42

Chúc bạn học tốt

\(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}\)và x+y=-60

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

\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)

=>x=-3.17=-51

y=-3.3=-9

b)\(\dfrac{x}{19}=\dfrac{y}{21}\Rightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)và 2x-y=34

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

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

=>x=2.19=38

y=2.21=42

\(\dfrac{x}{y}=\dfrac{17}{3} \)

\(\dfrac{x}{17}=\dfrac{y}{3}\)

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

\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)

\(\Rightarrow\)\(x=-3\Rightarrow 17 . -3 = -51 \)

\(\Rightarrow y =-3\Rightarrow 3 . -3=-9\)

a)

Theo tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}=\dfrac{y}{25}=\dfrac{x+y}{5+25}=\dfrac{60}{30}=2\)

\(\Rightarrow\dfrac{x}{5}=2\Rightarrow x=2\times5=10\)

\(\Rightarrow\dfrac{y}{25}=2\Rightarrow y=2\times25=50\)

Vậy\(\left\{{}\begin{matrix}x=10\\y=50\end{matrix}\right.\)

b)

\(\dfrac{x}{5}=\dfrac{y}{7}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{9}\right)^3\Rightarrow\dfrac{x}{5}\times\dfrac{x}{5}=\dfrac{x}{5}\times\dfrac{y}{7}=\dfrac{x\times y}{5\times7}=\dfrac{140}{35}=4=\left(2\right)^2\)

\(\Rightarrow\dfrac{x}{5}=2\Rightarrow x=2\times5=10\)

\(\Rightarrow\dfrac{y}{7}=2\Rightarrow y=2\times7=14\)

Vậy \(\left\{{}\begin{matrix}x=10\\y=14\end{matrix}\right.\)

1) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)

2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

⇒\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}\)

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

\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)

⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)

Đặt \(\dfrac{x}{-3}=\dfrac{y}{-8}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=8k\end{matrix}\right.\)

Ta có: \(x^2-y^2=-\dfrac{44}{5}\)

\(\Leftrightarrow9k^2-64k^2=-\dfrac{44}{5}\)

\(\Leftrightarrow k^2=\dfrac{4}{25}\)

Trường hợp 1: \(k=\dfrac{2}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=\dfrac{6}{5}\\y=8k=\dfrac{16}{5}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{2}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=\dfrac{-6}{5}\\y=8k=\dfrac{-16}{5}\end{matrix}\right.\)

\(\dfrac{x}{2}=\dfrac{y}{3}\) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}\) (1)

\(\dfrac{y}{4}=\dfrac{z}{5}\) ⇒ \(\dfrac{y}{12}=\dfrac{z}{15}\) (2)

Từ (1) và (2) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)\(=\dfrac{x+y-z}{8+12-15}\) \(=\dfrac{10}{5}=2\)

⇒ \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\\\dfrac{y}{12}=2\\\dfrac{z}{15}=2\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)

Ta có \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{1}{4}\cdot\dfrac{x}{2}=\dfrac{1}{4}\cdot\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{1}{3}\cdot\dfrac{y}{4}=\dfrac{1}{3}\cdot\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và x+y-z=10

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

\(\Rightarrow\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\)

\(\dfrac{y}{12}=2\Rightarrow=2\cdot12=24\)

\(\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\)

vậy x = 16; y = 24; z = 30

Chúc bn học tốt

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx+y-z=10\)

Ta có :

\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{4}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{5}\end{matrix}\right.\) => \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và \(x+y-z=10\)

Từ : \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}\)

=> \(\dfrac{10}{5}=2\)

Với : \(\dfrac{x}{8}=2\Rightarrow x=16\)

Với : \(\dfrac{y}{12}=2\Rightarrow y=24\)

Với: \(\dfrac{z}{15}=2\Rightarrow z=30\)

Vậy x,y,z là : 16,24,30

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{8}{3}}=\dfrac{x-y+z}{3-\dfrac{5}{2}+\dfrac{8}{3}}=\dfrac{95}{\dfrac{19}{6}}=30\\ \Rightarrow\left\{{}\begin{matrix}x=90\\y=30\cdot\dfrac{5}{2}=75\\z=30\cdot\dfrac{8}{3}=80\end{matrix}\right.\)

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

x+y=16 => x=16-y

thay vào đẳng thức đã cho, ta được:

\(\dfrac{3+16-y}{5+y}=\dfrac{3}{5}\Leftrightarrow\dfrac{19-y}{5+y}=\dfrac{3}{5}\\ \Leftrightarrow\left(19-y\right).5=3.\left(5+y\right)\\ \Leftrightarrow y=10\)

=> x = 6

vậy cặp số x,y cần tìm là 6;10

Theo bài ra ta có: x/3=y/4=z/6 và x-y+2z =121

\(\Rightarrow\)z/6 = 2z/12

\(\Rightarrow\)x/3=y/4=2z/12

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

\(\dfrac{x}{3}\)= \(\dfrac{y}{4}\)=\(\dfrac{2z}{12}\)= \(\dfrac{x-y+2z}{3-4+12}\)= \(\dfrac{121}{11}\)=11

+ \(\dfrac{x}{3}\)=11\(\Rightarrow\) x = 11 . 3 = 33

+ \(\dfrac{y}{4}\)=11 \(\Rightarrow\)y = 11.4 = 44

+\(\dfrac{2z}{12}\)=11 \(\Rightarrow\)2z = 11 . 12 =132

\(\Rightarrow\)z = 132 : 2 = 66

Vậy x = 33 ; y = 44 ; z = 66.