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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
\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{\left(6x-12y\right)+\left(8z-6x\right)+\left(12y-8z\right)}{9+4+16}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)
áp dụng tính chất dảy tỉ số bằng nhau
ta có : \(\dfrac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{\left(2.2\right)+\left(3.3\right)-4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}\)
\(=\dfrac{\left(2x+3y-z\right)-5}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)
suy ra ta có : \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=2.5\\y-2=3.5\\z-3=4.5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+1\\y=15+2\\z=20+3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\) vậy \(x=11;y=17;z=23\)
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 \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......
Từ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)
Và \(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\)\(\dfrac{y}{12}=\dfrac{z}{16}\)
Suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=-1\Rightarrow x=-1\cdot9=-9\\\dfrac{y}{12}=-1\Rightarrow y=-1\cdot12=-12\\\dfrac{z}{16}=-1\Rightarrow z=-1\cdot16=-16\end{matrix}\right.\)
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{y}{12}=\dfrac{z}{16}\)(2)
Từ (1) và (2) , suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ; ta được :
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)
Do đó :
\(\dfrac{x}{9}=-1\Rightarrow x=-1.9=-9\)
\(\dfrac{y}{12}=-1\Rightarrow y=-1.12=-12\)
\(\dfrac{z}{16}=-1\Rightarrow z=-1.16=-16\)
Vậy x = -9 ; y = -12 ; z = -16
Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{12.\left(x+y+z\right)}{49}\)
\(=\dfrac{12.49}{49}=12\)
\(\Rightarrow\dfrac{2x}{3}=12\Rightarrow x=18\)
\(\dfrac{3y}{4}=12\Rightarrow y=16\)
\(\dfrac{4z}{5}=12\Rightarrow z=15\)
Vậy \(x=18;y=16;z=15\)
Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)
⇒\(\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=12.\dfrac{3}{2}=18\)
⇒\(\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=12.\dfrac{4}{3}=16\)
⇒\(\dfrac{y}{\dfrac{5}{4}}=12\Rightarrow y=12.\dfrac{5}{4}=15\)
Vậy x;y;z lần lượt là 18;16;15
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{3\left(x-1\right)}{6}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\)
\(\Leftrightarrow\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x-3}{6}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{3x-3+3y-6-z+3}{6+9-4}=\dfrac{\left(3x+3y-z\right)+\left(3-3-6\right)}{11}=\dfrac{50-6}{11}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=4\Leftrightarrow x=4.2+1=9\\\dfrac{y-2}{3}=4\Leftrightarrow y=4.3+2=14\\\dfrac{z-3}{4}=4\Leftrightarrow z=4.4+3=19\end{matrix}\right.\)