2/3x=4/5y=5/6z và x+y+z=156
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{-5}=\frac{y}{6}=\frac{z}{-2}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow x=-5k;y=6k;z=-2k\)
\(\Rightarrow A=\frac{3.k.\left(-5\right)+6.k-2.\left(-2\right).k}{-3.\left(-5\right).k-5.6.k+6.\left(-2\right).k}=\frac{-15k+6k+4k}{15k-30k-12k}=\frac{-5k}{-27k}=\frac{5}{27}\)
Vậy \(A=\frac{5}{27}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=k\)\(\Rightarrow\begin{cases}x=5k+1\\y=3k+2\\z=2k+2\end{cases}\)
Theo đề bài: 3x-5y+6z <=> 3(5k+1)-5(3k+2)+6(2k+2)=9
<=>15k+3-15k-10+12k+12=9
<=>12k+5=9
<=>12k=4
<=>k=\(\frac{4}{12}=\frac{1}{3}\)
=>\(\Rightarrow\begin{cases}x=5.\frac{1}{3}+1=\frac{8}{3}\\y=3.\frac{1}{3}+2=3\\z=2.\frac{1}{3}+2=\frac{8}{3}\end{cases}\)
Vậy ............
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}\) = \(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\)
= \(\frac{3x-3-\left(5y-10\right)+6z-12}{15-15+12}\) = \(\frac{3x-3-5y+10+6x-12}{12}\)
= \(\frac{9-5}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)
=> \(\left[\begin{array}{nghiempt}x-1=\frac{5}{3}\\y-2=1\\z-2=\frac{2}{3}\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x=\frac{8}{3}\\y=3\\z=\frac{8}{3}\end{array}\right.\)
Vậy x = \(\frac{8}{3}\) ; y = 3 ; z = \(\frac{8}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}=>\frac{y}{15}=\frac{z}{21}\)
=>\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
=> x=2x10=20
y=2x15=30
z=2x21=42
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )
5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)
\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)
b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)
\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(\frac{x-1}{5}\) = \(\frac{y-2}{3}\) = \(\frac{z-2}{2}\) => \(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\) và 3x-5y+6z =9
Áp dụng t/c ..., ta có:
\(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\) =\(\frac{\left(3x-5y+6z\right)+\left(-3+10-12\right)}{15-15+12}\) =\(\frac{4}{12}\)=\(\frac{1}{3}\)
\(\frac{x-1}{5}\) =\(\frac{1}{3}\) =>x-1=\(\frac{5}{3}\)=>x=\(\frac{8}{3}\)
\(\frac{y-2}{3}\) = \(\frac{1}{3}\)=>y-2=1 =>y=3
\(\frac{z-2}{2}\) =\(\frac{1}{3}\) =>z-2=\(\frac{2}{3}\) =>z=\(\frac{8}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Theo đề ra, ta có:
\(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{6}\)
\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{6}=\frac{z}{5}\)
Áp dụng tính chất của dãy tỷ số bằng nhau
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{5}=\frac{x-y}{10-6}=1\)
\(\Rightarrow x=1.10=10\)
\(\Rightarrow y=1.6\)
\(\Rightarrow z=1.5=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 3x = 5y = 6z
=> \(\frac{x}{5}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)
\(=>\frac{x}{30}=\frac{y}{18};\frac{y}{18}=\frac{z}{15}\)
\(hay\)\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}=\frac{x+y-z}{30+18-15}=\frac{22}{33}=\frac{2}{3}\)
Do đó suy ra:
\(3x=\frac{2}{3}=>x=\frac{2}{9}\)
\(5y=\frac{2}{3}=>y=\frac{2}{15}\)
\(6z=\frac{2}{3}=>x=\frac{1}{9}\)
Vậy \(\left(x;y;z\right)\in\left\{\frac{2}{9};\frac{2}{15};\frac{1}{9}\right\}\)
Có: \(\frac{2}{3x}=\frac{4}{5y}=\frac{5}{6z}\Rightarrow\frac{3x}{\frac{1}{2}}=\frac{5y}{\frac{1}{4}}=\frac{6z}{\frac{1}{5}}\Rightarrow\frac{x}{\frac{1}{6}}=\frac{y}{\frac{1}{20}}=\frac{z}{\frac{1}{30}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{\frac{1}{6}}=\frac{y}{\frac{1}{20}}=\frac{z}{\frac{1}{30}}=\frac{x+y+z}{\frac{1}{6}+\frac{1}{20}+\frac{1}{30}}=\frac{156}{\frac{1}{4}}=624\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{\frac{1}{6}}=624\Rightarrow x=624\cdot\frac{1}{6}=104\\\frac{y}{\frac{1}{20}}=624\Rightarrow y=624\cdot\frac{1}{20}=31,2\\\frac{z}{\frac{1}{30}}=624\Rightarrow z=624\cdot\frac{1}{30}=20,8\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(104;31,2;20,8\right)\)