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a) \(\frac{-3}{5}.y=\frac{21}{10}\)
\(y=\frac{21}{10}:\frac{-3}{5}\)
\(y=\frac{-7}{2}\)
vậy \(y=\frac{-7}{2}\)
b) \(y:\frac{3}{8}=-1\frac{31}{33}\)
\(y:\frac{3}{8}=\frac{-64}{33}\)
\(y=\frac{-64}{33}.\frac{3}{8}\)
\(y=\frac{-8}{11}\)
vậy \(y=\frac{-8}{11}\)
c) \(1\frac{2}{5}.y+\frac{3}{7}=\frac{-4}{5}\)
\(\frac{7}{5}.y+\frac{3}{7}=\frac{-4}{5}\)
\(\frac{7}{5}.y=\frac{-4}{5}-\frac{3}{7}\)
\(\frac{7}{5}.y=\frac{-43}{35}\)
\(y=\frac{-43}{35}:\frac{7}{5}\)
\(y=\frac{-43}{49}\)
vậy \(y=\frac{-43}{49}\)
d) \(\frac{-11}{12}.y+0,25=\frac{5}{6}\)
\(\frac{-11}{12}.y=\frac{5}{6}-0,25\)
\(\frac{-11}{12}.y=\frac{7}{12}\)
\(y=\frac{7}{12}:\frac{-11}{12}\)
\(y=\frac{-7}{11}\)
vậy \(y=\frac{-7}{11}\)
a) \(y=\frac{21}{10}:\left(-\frac{3}{5}\right)=-\frac{7}{2}\)
b) \(y=-1\frac{31}{33}.\frac{3}{8}=-\frac{8}{11}\)
c) \(1\frac{2}{5}.y=-\frac{4}{5}-\frac{3}{7}=-\frac{43}{35}\Rightarrow y=-\frac{43}{35}:1\frac{2}{5}=-\frac{43}{49}\)
d) \(-\frac{11}{12}.y=\frac{5}{6}-0,25=\frac{7}{12}\Rightarrow y=\frac{7}{12}:\left(-\frac{11}{12}\right)=-\frac{7}{11}\)
-11/12.y+1/4=5/6
-11/12.y =5/6-1/4
-11/12.y =7/12
y =7/12:-11/12
y =-7/11
\(\frac{-11}{12}.y+0,25=\frac{5}{6};-\frac{11}{12}.y=\frac{7}{12};y=-\frac{7}{11}\)
=> 12(111 - x) = 11(120 - x)
12.111 - 12x = 11.120 - 11.x
1332 - 12x = 1320 - 11x
1332 - 1320 = 12x - 11x
12 = x
=> x = 12
Có \(\frac{120-12}{12}=\frac{y}{6}\)
=> 9 = \(\frac{y}{6}\)
y = 9.6
y = 54
Vậy x = 12; y = 54
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{111-x}{11}=\frac{120-x}{12}=\frac{y}{6}=\frac{120-x-111+x}{12-11}=\frac{9}{1}=9\)
\(\Rightarrow\hept{\begin{cases}\frac{111-x}{11}=9\\\frac{y}{6}=9\end{cases}\Rightarrow\hept{\begin{cases}111-x=9.11=99\\y=9.6=54\end{cases}\Rightarrow}\hept{\begin{cases}x=12\\y=54\end{cases}}}\)
Vậy \(x=12;y=54\)
b) Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow x=12k,y=9k,z=5k\)
\(xyz=20\)
\(\Rightarrow12k.9k.5k=20\)
\(\Rightarrow540k^3=20\)
\(\Rightarrow k^3=\frac{1}{27}\)
\(\Rightarrow k=\frac{1}{3}\)
Khi \(k=\frac{1}{3}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{3}\Rightarrow x=4\)
\(\frac{y}{9}=\frac{1}{3}\Rightarrow y=3\)
\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=\frac{5}{3}\)
Vậy x = ..... ; y = ............ ; z = .............
a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)
=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)
\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(48;12\right).\)
b)
\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
⇒ \(\frac{5}{3}x=\frac{1}{21}\)
⇒ \(x=\frac{1}{21}:\frac{5}{3}\)
⇒ \(x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}.\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
⇒ \(x-\frac{1}{2}=\frac{1}{3}\)
⇒ \(x=\frac{1}{3}+\frac{1}{2}\)
⇒ \(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}.\)
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a)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\)x/12=4 suy ra x=12.4=48
y/3=4 suy ra y=3.4 =12
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=2\)
suy ra x-2/5=2 hoac x-2/5=-2
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
\(\frac{-11}{12}y+0,25=\frac{5}{6}\)
\(\frac{-11}{12}y+\frac{1}{4}=\frac{5}{6}\)
\(\frac{-11}{12}y=\frac{5}{6}-\frac{1}{4}\)
\(\frac{-11}{12}y=\frac{7}{12}\)
\(y=\frac{7}{12}:\frac{-11}{12}\)
\(y=\frac{-7}{11}\)