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\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chát dãy tỉ số = nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=3\Rightarrow z=63\)
b, Tự làm
c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)
\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)
\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)
\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)
Vậy \((x,y)\in(6,15);(-6,-15)\)
4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
a) Từ x:y:z = 3:5:(-2) => \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{124}{4}=31\)
=> \(\begin{cases}x=93\\y=155\\z=-62\end{cases}\)
b) Từ \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\)
\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3z-7y+5z}{63-98+50}=\frac{30}{15}=2\)
=> \(\begin{cases}x=42\\y=28\\z=20\end{cases}\)
a) Giải:
Ta có: \(x:y:z=3:5:\left(-2\right)\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x}{15}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{124}{4}=31\)
+) \(\frac{x}{3}=31\Rightarrow x=93\)
+) \(\frac{y}{5}=31\Rightarrow y=155\)
+) \(\frac{z}{-2}=31\Rightarrow z=-62\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(93;155;-62\right)\)
b) Giải:
Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\)
\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
+) \(\frac{x}{21}=2\Rightarrow x=42\)
+) \(\frac{y}{14}=2\Rightarrow y=28\)
+) \(\frac{z}{10}=2\Rightarrow z=20\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(42;28;20\right)\)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
Trả lời:
1, Ta có: \(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{4}\)
\(\Rightarrow x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Rightarrow2x+2y+2z=\frac{13}{12}\)
\(\Rightarrow2\left(x+y+z\right)=\frac{13}{12}\)
\(\Rightarrow x+y+z=\frac{13}{24}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\\y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\\z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\end{cases}}\)
2, Ta có: \(x:y:z=3:5:\left(-2\right)\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)
Áp dụng tc dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x-y+3z}{5.3-5+3.\left(-2\right)}=\frac{124}{4}=31\)
\(\Rightarrow\hept{\begin{cases}x=93\\y=155\\z=-62\end{cases}}\)
3, Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\left(1\right)\)
\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
Áp dụng tc dãy tỉ số bằng nhau, ta có:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5x}{3.21-7.14+5.10}=\frac{30}{15}=2\)
\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)
a) Ta có : 2x = 3y => \(\frac{x}{3}=\frac{y}{2}\)
7z = 5y => \(\frac{y}{7}=\frac{z}{5}\)
=> \(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)
+) \(\frac{x}{3}=\frac{y}{2}\)=> \(\frac{x}{21}=\frac{y}{14}\)
+) \(\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
=> x = 2.21 = 42 , y = 2.14 = 28 , z = 2.10 = 20
b) Ta có : x : y : z = 3 : 5 : (-2) => \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)
Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\\z=-2k\end{cases}}\)
=> 5x = 15k , y = 5k , 3z = -6k
=> 5x - y + 3z = 15k - 5k + (-6k)
=> -16 = 10k - 6k
=> -16 = 4k
=> k = -4
Với k = -4 thì x = 3.(-4) = -12 , y = 5.(-4) = -20 , z = (-2).(-4) = 8
Vậy : ....
minh lam cau b) roi dc co 2/3 thoy ban tham khao nhe phan () la minh giai thich nha dung viet vo bai !!
2x=3y ; 5y = 7z
+) 10x=15y=21z ( Quy dong)
+)10x/210 = 15y/210 = 21z/210 ( BC)
+) x/21 = y/14 = z/10 ( Rut gon)
+) 3x/63 = 7y/98 = 5z/50 = 3x-7y+ 5z / 63 - 98 - 50 = -30/14 = -2
+ x/21 = 2 => ............ phan nay minh chua xong neu xong thi minh pm not cho
Câu c là dấu " . " là dấu nhân
a) \(x:y:z=3:5:\left(-2\right)\) => \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)=> \(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}\)
Áp dụng TC dãy tỉ số bằng nhau ta có ;
\(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{124}{4}=31\)
=> \(\hept{\begin{cases}\frac{x}{3}=31\\\frac{y}{5}=31\\\frac{z}{-2}=31\end{cases}}\Rightarrow\hept{\begin{cases}x=93\\y=155\\z=-62\end{cases}}\)
b) Ta có : \(\hept{\begin{cases}2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\\5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\end{cases}}\)
=> \(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)
=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
=> \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{-30}{15}=-2\)
=> \(\hept{\begin{cases}\frac{x}{21}=-2\\\frac{y}{14}=-2\\\frac{z}{10}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=-42\\y=-28\\z=-20\end{cases}}\)
c) Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
=> \(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
=> xyz = 2k.3k.5k
=> 30k3 = 810
=> k3 = 27
=> k = 3
Vậy x = 6,y = 9,z = 15