Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có:
\(x+y+z=49\Rightarrow12x+12y+12z=588\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{588}{49}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.3:2\\y=12.4:3\\z=12.5:4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
⇒2x = 3.30 = 90 ⇒ x = 45
3y = 3.60 = 180 ⇒ y = 60
z = 3.28 = 84
Ý b) có gì đó sai sai ?
c)Ta có :
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
⇒x = 5.15 = 75
y = 5.10 = 50
z = 5.6 = 30
d)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)
⇒ x = 2k ; y = 3k ; z = 5k
⇒ xyz = 2k.3k.5k = 30k3 = 810
⇒ k = 3 Vậy x = 3.2 = 6; y = 3.3 = 9; z = 3.5 = 15
a,3x=2y;7y=5z
=>\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta co:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\\ \Rightarrow x=2.10=20\\ y=2.15=30\\ z=2.21=42\)
Các câu sau tương tự
b,\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\),\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\) và 2x-3y+z=6
Từ đề bài ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)\(\Rightarrow\)\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)(2)
từ (1) và (2)\(\Rightarrow\)\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{20}\)\(\Rightarrow\)\(\dfrac{2x}{18}\)=\(\dfrac{3y}{36}\)=\(\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x}{18}\)=\(\dfrac{3y}{36}\)=\(\dfrac{z}{20}\)=\(\dfrac{2x-3y+z}{18-36+20}\)=\(\dfrac{6}{2}\)=3
\(\Rightarrow\)x=3.9=27
y=3.12=36
z=3.20=60
Vậy.....
chúc bạn học tốt,nhớ tick cho mình nha
d) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(xyz=810\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
=> \(x=2k\) ; \(y=3k\) ; \(z=5k\)
Thay \(x=2k;y=3k;z=5k\) vào \(xyz=810\) ta được
\(2k.3k.5k=810\)
\(30k=810\)
\(k^3=27\)
=> k = 3
=> \(x=2.3=6\)
=> \(y=3.3=9\)
=> \(z=5.3=15\)
a) Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\cdot\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\)
\(\Rightarrow\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)
\(\Rightarrow\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\)
\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\)
Thay vào \(y+z+1=2x\) ; ta có :
\(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\)
Thay vào \(x+z+2=2y\) ; ta có :
\(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+y=\dfrac{1}{2}-z\)
Thay vào \(x+y-3=2z\) ; ta có :
\(\dfrac{1}{2}-z-3=2z\Rightarrow3z=\dfrac{-5}{2}\Rightarrow z=\dfrac{-5}{6}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
1, \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)\(\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\)\(\Leftrightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)
Mà xyz = -108
\(\Leftrightarrow2k.\frac{3}{2}k.\frac{4}{3}k=-108\)
\(\Leftrightarrow4k^3=-108\)
<=> k3 = -27
<=> k = -3
\(\Leftrightarrow\hept{\begin{cases}x=2k=2.-3=-6\\y=\frac{3}{2}k=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}k=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)
2, \(\frac{x}{5}=\frac{y}{7}=\frac{z}{8}\)\(\Leftrightarrow\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}=\frac{2x+3y-4z}{10+21-32}=\frac{15}{-1}=-15\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-15\\\frac{y}{7}=-15\\\frac{z}{8}=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=-75\\y=-105\\z=-120\end{cases}}\)
3, 3x = 5y \(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)\(\Leftrightarrow\frac{x}{55}=\frac{y}{33}\)
2y = 11z \(\Leftrightarrow\frac{y}{11}=\frac{z}{2}\) \(\Leftrightarrow\frac{y}{33}=\frac{z}{6}\)
\(\Rightarrow\frac{x}{55}=\frac{y}{33}=\frac{z}{6}\)\(\Rightarrow\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}=\frac{2x+5y-z}{110+165-6}=\frac{34}{269}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{55}=\frac{34}{269}\\\frac{y}{33}=\frac{34}{269}\\\frac{z}{6}=\frac{34}{269}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1870}{269}\\y=\frac{1122}{269}\\z=\frac{204}{269}\end{cases}}\)
4, \(\frac{x}{3}=\frac{2}{y}=\frac{z}{4}=k\)\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=\frac{2}{k}\\z=4k\end{cases}}\)
Mà xyz = 240
<=> 3k . 2/k . 4k = 240
<=> 24k = 240
<=> k = 10
\(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)
\(x-y+100=z\Rightarrow x-y-z=-100\)
\(\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{y}{15}=\dfrac{z}{9}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
\(\Rightarrow x=20.25=500;y=15.25=375;z=9.25=225\)
b/ \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
\(\Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{5z-25-4y-12-3x+3}{30-16-6}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=2\\\dfrac{y+3}{4}=2\\\dfrac{z-5}{6}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
c/ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=a\Rightarrow\left\{{}\begin{matrix}x=2a\\y=3a\\z=5a\end{matrix}\right.\) \(\Rightarrow xyz=2a.3a.5a=30a^3=-30\Rightarrow a^3=-1\Rightarrow a=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2a=-2\\y=3a=-3\\z=5a=-5\end{matrix}\right.\)
d/ \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\Rightarrow\dfrac{2x}{2,2}=\dfrac{y}{1,3}=\dfrac{z}{1,4}=\dfrac{2x-y}{2,2-1,3}=\dfrac{5,5}{0,9}=\dfrac{55}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1,1.55}{9}=\dfrac{121}{18}\\y=\dfrac{1,3.55}{9}=\dfrac{143}{18}\\z=\dfrac{1,4.55}{9}=\dfrac{77}{9}\end{matrix}\right.\) Nghi ngờ bạn chép đề câu này sai, số xấu quá
a)Ta có: \(\frac{x}{y+z+1}=\frac{y}{x+y+2}=\frac{z}{x+y-3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{y+z+1}=\frac{y}{x+y+2}=\frac{z}{x+y-3}\)
\(=\frac{x+y+z}{y+z+1+x+y+2+x+y-3}\)
\(=\frac{x+y+z}{2x+2y+2z}\)
\(=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
Ta có : \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}\left(1\right)\)
\(5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\Rightarrow\dfrac{y}{18}=\dfrac{z}{15}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}=k\Rightarrow x=24k;y=18k;z=15k\)
\(\text{Ta có }:x.y.z=24k.18k.15k=30\\ \Rightarrow k^3.6480=30\\ \Rightarrow k^3=\dfrac{1}{216}\\ \Rightarrow k=\dfrac{1}{6}\\ \Rightarrow x=24.\dfrac{1}{6}=4\\ y=18.\dfrac{1}{6}=3\\ z=15.\dfrac{1}{6}=2.5\)
Vậy x = 4 ; y = 3 ; z = 2,5
a) Tìm các số x, y, z biết rằng
3x=4y ; 5y=6z và xyz = 30
Giải
Ta có: 3x = 4y; 5y = 6z \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\); \(\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Có \(xyz=30\) \(\Leftrightarrow\) \(8k.6k.5k=30\)
\(\Rightarrow\) \(240k^3=30\)
\(\Rightarrow\) \(k^3=8\)
\(\Rightarrow\) \(k=\sqrt[3]{8}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)
a) \(2x=5y\)⇒\(x=\dfrac{5}{2}y\)⇒\(xy=\dfrac{5}{2}y^2\)
Thay \(xy=250\), ta có:
\(250=\dfrac{5}{2}y^2\)
⇒\(y^2=100\)⇒\(y=+-10\)
+) \(y=10\text{⇒}x=250:10=25\)
+) \(y=-10\text{⇒}x=250:-10=-25\)
\(a,2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=k\\ \Rightarrow x=5k;y=2k\\ xy=250\Rightarrow5k\cdot2k=250\Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=25;y=10\\x=-25;y=-10\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=a\Rightarrow x=3a;y=2a;z=4a\\ xyz=192\Rightarrow24a^3=192\Rightarrow a^3=8\Rightarrow a=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=4\\z=8\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=q\Rightarrow x=5q;y=2q;z=-3q\\ xyz=240\Rightarrow-30q^3=240\Rightarrow q^3=-8\Rightarrow q=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-10\\y=-4\\z=6\end{matrix}\right.\)