tìm x biết : x\(\frac{x}{3}\)= 24
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A ) ĐK: x#0
Ta có:
(1) 1+2y/18 = 1+4y/24
=> 24 + 48y = 18 + 72y
<=> y=1/4
(2) 1+4y/24=1+6y/6x
Thay y=1/4 vào (2) ta tìm đc x=5 (thỏa)
B ) x+y=3(x−y)=x:y
→x+y=3x−3y
→4y=2x
→x:y=4:2=2
→x+y=2
Mà x=2y nên
2y+y=3y=2
→y=2/3
→x=2−2/3=4/3
Chú ý : dấu / nghĩa là phần
Nếu mình đúng thì các bạn k mình nhé
a) \(\frac{1+2y}{18}=\frac{1+4y}{24}\Rightarrow24+48y=18+72y\Rightarrow6=24y\Rightarrow y=\frac{1}{4}\)
\(\frac{1+4y}{24}=\frac{1+6y}{6x}\Rightarrow\frac{1+4.\frac{1}{4}}{24}=\frac{1+6.\frac{1}{4}}{6x}\Rightarrow\frac{2}{24}=\frac{\frac{5}{2}}{6x}\Rightarrow12x=60\Rightarrow x=5\)
b) \(x+y=3\left(x-y\right)\Rightarrow x+y=3x-3y\Rightarrow4y=2x\Rightarrow x=2y\)
\(x+y=\frac{x}{y}\Rightarrow2y+y=\frac{2y}{y}\Rightarrow3y=2\Rightarrow y=\frac{2}{3}\Rightarrow x=2y=\frac{4}{3}\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(\dfrac{2}{3}x-1\right)^3\cdot\dfrac{1}{50}=\dfrac{75}{2}-24=\dfrac{27}{2}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-1\right)^3=\dfrac{27}{2}:\dfrac{1}{50}=675\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}-1=3\sqrt[3]{25}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=3\sqrt[3]{25}+1\)
\(\Leftrightarrow x=\dfrac{3\left(3\sqrt[3]{25}+1\right)}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1-\(\left(3\frac{3}{8}+x-2\frac{5}{24}\right)\)\(.\frac{12}{17}=0\)
\(\Rightarrow\left(3\frac{3}{8}+x-2\frac{5}{24}\right)\)\(.\frac{12}{17}\)\(=1\)
\(\frac{27}{8}+x-\frac{53}{24}\) \(=1:\frac{12}{17}\)
\(\frac{27}{8}+x=\frac{17}{12}+\frac{53}{24}\)
\(\frac{27}{8}+x=\frac{34+53}{24}\)
\(\frac{27}{8}+x=\frac{87}{24}\)
\(x=\frac{87}{24}-\frac{27}{8}\)
\(x=\frac{87-81}{24}\)
\(x=\frac{6}{24}=\frac{1}{4}\)
<=> x^2/3=24
<=> x^2 = 24*3
<=> x^2=72
<=> x=\(\sqrt{72}\)
mk ko ghi de nha chuc bn hoc tot