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a: x=2/3-4/5=10/15-12/15=-2/15
b: 1/2-x=7/12
=>x=1/2-7/12=-1/12
c: =>7/2:x=-7/2
=>x=-1
d: =>1/6x=3/8-5/2=3/8-20/8=-17/8
=>x=-17/8*6=-102/8=-51/4
e: =>1,5x=-1,5
=>x=-1
\(2^x.4^2-2^{x+1}=2^6-2^3\)
\(2^x.2^4-2^x.2=2^2.\left(2^4-2\right)\)
\(2^x.\left(2^4-2\right)=2^2.\left(2^4-2\right)\)
\(2^x.14=2^2.14\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3^{50}=\left(3^5\right)^{10}=243^{10}\)
\(5^{30}=\left(5^3\right)^{10}=125^{10}\)
Ta có: \(125< 243\)
\(\Rightarrow125^{10}< 243^{10}\)
\(\Rightarrow3^{50}< 5^{30}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a) x/3 - 1/4 = - 5/6
x/3 = -5/6 + 1/4
x/3 = -7/12
x = -7/12 :3
x = -7/36