\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và x+y+z
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\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-6x}{-11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)
\(\Rightarrow\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)
\(-x=90\times\frac{-11}{6}=-165\Rightarrow x=165\)
\(y=90\times\frac{2}{9}=20\)
\(z=90\times\frac{5}{18}=25\)
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\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
=>x=165,y=20,z=25
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a, \(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\Rightarrow\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2\)
=> x=8,y=6,z=18
b, \(\hept{\begin{cases}\frac{x}{y}=\frac{9}{7}\Rightarrow\frac{x}{9}=\frac{y}{7}\\\frac{y}{z}=\frac{7}{3}\Rightarrow\frac{y}{7}=\frac{z}{3}\end{cases}\Rightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3}\)
=> x=-27,y=-21,z=-9
c, \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\Rightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
=> x=165,y=20,z=25
Ba số x, y, z thỏa mãn :\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và -x + y + z = -120
Hỏi x+y+z=?
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a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)
8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)
=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)
=> x = 24,y = 15,z = 6
b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)
\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)
=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)
=> x = -165 , y = -20 , z = -25
c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k
=> xyz = 12k . 9k . 5k
=> xyz = 540k3
=> 540k3 =20
=> k3 = 20/540
=> k3 = 1/27
=> k = 1/3
Do đó : x= 4 , y = 3 , z = 5/3
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\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{-\frac{11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{-\frac{4}{3}}=90\)
\(\frac{x}{\frac{11}{6}}=90\Rightarrow x=90\times\frac{11}{6}=165\)
\(\frac{y}{\frac{2}{9}}=90\Rightarrow y=90\times\frac{2}{9}=20\)
\(\frac{z}{\frac{5}{18}}=90\Rightarrow x=90\times\frac{5}{18}=25\)
Giải:
Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)
+) \(\frac{x}{\frac{11}{6}}=90\Rightarrow x=165\)
+) \(\frac{y}{\frac{2}{9}}=90\Rightarrow y=20\)
+) \(\frac{z}{\frac{5}{18}}=20\Rightarrow z=25\)
Vậy bộ số \(\left(x,y,z\right)\) là: \(\left(165,20,25\right)\)
Tự nhiên máy mk bị restart nên mk gửi trả lời hơi chậm nhé!