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
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
a)tìm x,y biết:
Ta có : \(\dfrac{x}{12} = \dfrac{y}{3}\) \(\Rightarrow\) \(\dfrac{x-y}{12-3} = \dfrac{36}{9} = 4\)
b)tìm x biết:
23+53x=57
\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{5}{7} - \dfrac{2}{3}\)
\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{1}{21}\)
\(\Leftrightarrow\) \(x = \dfrac{1}{35}\)
1112−(25+x)=23
\(\Leftrightarrow\) \(\dfrac{2}{5} +x\) = \(\dfrac{2}{3} - \dfrac{11}{12}\)
\(\Leftrightarrow\) \(\dfrac{2}{5} + x = \dfrac{-1}{4}\)
\(\Leftrightarrow\) \(x = \dfrac{-13}{20}\)
|x−25|+34=114
\(\Leftrightarrow\) \(x - \dfrac{2}{5} = 2\)
\(\Leftrightarrow\) \(x = \dfrac{8}{5}\)
\(\Leftrightarrow\) \(x - \dfrac{1}{2} = \dfrac{1}{19683}\)
\(\Leftrightarrow\) \(x = \dfrac{19685}{39366}\)
a)\(\Rightarrow x=4y\)
Thay vào,ta đc:
\(3y=36\Leftrightarrow y=12\Leftrightarrow x=48\)
Vậy (x;y)=(48;12).
b) *\(\Leftrightarrow x=\frac{\frac{5}{7}-\frac{2}{3}}{\frac{5}{3}}=\frac{1}{35}\)
Vậy x=\(\frac{1}{35}\)
Ttự ta đc: *(2):\(x=\frac{5}{6}\)
*(3) cx thế
#Walker