Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
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
4x = 3y => x/3 = y/4 => x/9 = y/12 ( 1 )
5y = 6z => y/6 = z/5 => y/12 = z/10 ( 2 )
Từ ( 1 ) và ( 2 ) => x/9 = y/12 = z/10
=> 2x/18 = y/12 = z/10
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
2x/18 = y/12 = z/10 = 2x+y-z/18+12-10 = 40/20 = 2
=> x = 18 ; y = 24 ; z = 20
Vậy ...
Ta có :\(15x=10y=6z\Rightarrow\hept{\begin{cases}15x=10y\\10y=6z\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Khi đó 5x3 + 2y3 - z3 = 31
=> 5(2k)3 + 2(3k)3 - (5k)3 = 31
=> 40k3 + 54k3 - 125k3 = 31
=> -31k3 = 31
=> k3 = -1
=> k = -1
=> x = -2 ; y = -3 ; z = -5
b) Ta có 7x = 14y = 6z => \(\hept{\begin{cases}7x=14y\\14y=6z\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\7y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{1}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{6}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\frac{x}{6}=\frac{y}{3}=\frac{z}{7}\)
Đặt \(\frac{x}{6}=\frac{y}{3}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=6k\\y=3k\\z=7k\end{cases}}\)
Khi đó 2x2 - 3y2 = 5
<=> 2.(6k)2 - 3.(3k)2 = 5
=> 72k2 - 27k2 = 5
=> 45k2 = 5
=> k2 = 1/9
=> k = \(\pm\frac{1}{3}\)
Nếu k = 1/3 => x = 2 ; y = 1 ; z = 7/3
Nếu k = -1/3 => x = -2 ; y = - 1 ; z = -7/3
Vậy các cặp (x;y;z) thỏa mãn là : (2;1;7/3) ; (-2 ; - 1; -7/3)
c) Ta có : \(3x=8y=5z\Rightarrow\frac{3x}{120}=\frac{8y}{120}=\frac{5z}{120}\Rightarrow\frac{x}{40}=\frac{y}{15}=\frac{z}{24}\)
Đặt \(\frac{x}{40}=\frac{y}{15}=\frac{z}{24}=k\Rightarrow\hept{\begin{cases}x=40k\\y=15k\\z=24k\end{cases}}\)
Khi đó |x - 2y| = 5
<=> |40k - 2.15k| = 5
=> |10k| = 5
=> \(\orbr{\begin{cases}10k=5\\10k=-5\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)
Nếu k = 5 => x = 20 ; y = 7,5 ; z = 12
Nếu k = -5 => x = -20 ; y =-7,5 ; z = -12
d) 4x = 5y = 6z => \(\frac{4x}{60}=\frac{5y}{60}=\frac{6z}{60}\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{10}\)
Đặt \(\frac{x}{15}=\frac{y}{12}=\frac{z}{10}=k\Rightarrow\hept{\begin{cases}x=15k\\y=12k\\z=10k\end{cases}}\)
Khi đó (3x - 2y)2 = 16
<=> (3.15k - 2.12k)2 = 16
=> (45k -24k)2 = 16
=> (21k)2 = 16
=> \(\orbr{\begin{cases}21k=4\\21k=-4\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{4}{21}\\k=-\frac{4}{21}\end{cases}}\)
Nếu k = 4/21 => x = 20/7 ; y = 16/7 ; z = 40/21
Nếu k = -4/21 => x = -20/7 ; y = -16/7 ; z = -40/21
Vì \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\)
\(4x=6z\Rightarrow\frac{x}{6}=\frac{z}{4}\Rightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\)
\(\Rightarrow\frac{2x}{6}=\frac{7y}{28}=\frac{3z}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{6}=\frac{7y}{28}=\frac{3z}{6}=\frac{2x+7y-3z}{6+28-6}=\frac{2}{28}=\frac{1}{14}\)
\(\cdot\frac{x}{3}=\frac{1}{14}\Rightarrow x=\frac{3}{14}\)
\(\cdot\frac{y}{4}=\frac{1}{14}\Rightarrow y=\frac{2}{7}\)
\(\cdot\frac{z}{2}=\frac{1}{14}\Rightarrow z=\frac{1}{7}\)
a) \(\frac{2x}{3}=\frac{3y}{4}\Leftrightarrow8x=9y\Rightarrow x=\frac{9y}{8}\left(1\right)\)
\(\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow15y=16z\Rightarrow z=\frac{15y}{16}\left(2\right)\)
THay (1) và (2) vào biểu thức \(x+y+z=41\);ta được : \(\frac{9y}{8}+y+\frac{15y}{16}=41\)
\(\Rightarrow18y+16y+15y=656\Rightarrow y=\frac{656}{49}\)
Do đó : \(x=\frac{\frac{9.656}{49}}{8}=\frac{738}{49}\)
\(z=\frac{\frac{15.656}{49}}{16}=\frac{615}{49}\)
KL : \(x=\frac{738}{49};y=\frac{656}{49};z=\frac{615}{49}\)
b) Ta có : \(4x=3y\Rightarrow x=\frac{3y}{4}\)(1)
\(5y=6z\Rightarrow z=\frac{5y}{6}\)(2)
Thay (1) và (2) vào biểu thức \(x^2+y^2+z^2=500\);ta được :
\(\left(\frac{3y}{4}\right)^2+y^2+\left(\frac{5y}{6}\right)^2=500\)
\(\Rightarrow\frac{9y^2}{16}+y^2+\frac{25y^2}{36}=500\Rightarrow324y^2+576y^2+400y^2=288000\)
\(\Rightarrow1300y^2=288000\Rightarrow y^2=\frac{2880}{13}\Rightarrow\orbr{\begin{cases}y=\frac{24\sqrt{65}}{13}\\y=-\frac{24\sqrt{65}}{13}\end{cases}}\)
Với \(y=\frac{24\sqrt{65}}{13}\Rightarrow x=\frac{3\cdot\frac{24\sqrt{65}}{13}}{4}=\frac{18\sqrt{65}}{13};z=\frac{5\cdot\frac{24\sqrt{65}}{13}}{6}\)
\(y=-\frac{24\sqrt{65}}{13}\Rightarrow x=-\frac{18\sqrt{65}}{13};z=\frac{5\cdot-\frac{24\sqrt{65}}{13}}{6}\)
Giải:
Ta có:
\(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)
\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)
Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=k\)
\(\Rightarrow x=9k,y=12k,z=10k\)
Ta có:
\(A=\frac{2x^2-3y^2-4z^2}{4xy-3yz+2xz}\)
\(\Rightarrow A=\frac{2\left(9k\right)^2-3\left(12k\right)^2-4\left(10k\right)^2}{4.9.k.12.k-3.12.k.10.k+2.9.k.10k}\)
\(\Rightarrow A=\frac{2.9.k^2-3.12.k^2-4.10.k^2}{432.k^2-360.k^2+180.k^2}\)
\(\Rightarrow A=\frac{18.k^2-36.k^2-40.k^2}{k^2.\left(432-360+180\right)}\)
\(\Rightarrow A=\frac{k^2.\left(18-36-40\right)}{k^2.252}\)
\(\Rightarrow A=\frac{-58}{252}\)
\(\Rightarrow A=\frac{-1}{3}\)
Vậy \(A=\frac{-1}{3}\)
Ta có:
\(\begin{cases}4x=3y\\5y=6z\end{cases}\) => \(\begin{cases}\frac{x}{3}=\frac{y}{4}\\\frac{y}{6}=\frac{z}{5}\end{cases}\)=> \(\begin{cases}\frac{x}{9}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{10}\end{cases}\) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)
Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=k\)
=> \(\begin{cases}x=9k\\y=12k\\z=10k\end{cases}\)
Ta có:
\(A=\frac{2.\left(9k\right)^2-3.\left(12k\right)^2-4.\left(10k\right)^2}{4.9k.12k-3.12k.10k+2.9k.10k}\)
\(A=\frac{2.81.k^2-3.144.k^2-4.100.k^2}{432k^2-360k^2+180k^2}\)
\(A=\frac{162k^2-432k^2-400k^2}{252k^2}\)
\(A=\frac{-670k^2}{252k^2}=\frac{-335}{126}\)