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8 tháng 6 2017
\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}.....\dfrac{30}{62}.\dfrac{31}{64}=2x\)
\(\dfrac{1.2.3.4.5.....30.31}{\left(2.2\right)\left(2.3\right)\left(2.4\right)\left(2.5\right)\left(2.6\right).....\left(2.31\right)\left(2.32\right)}=2x\)
\(\dfrac{1\left(2.3.4.5....30.31\right)}{32\left(2.3.4.5.....31\right).2^{31}}=2x\)
\(\dfrac{1}{2^5.2^{31}}=2x\Rightarrow2x=\dfrac{1}{2^{36}}\Rightarrow x=\dfrac{1}{2^{36}}\div2=\dfrac{1}{2^{37}}\)
Vậy x = \(\dfrac{1}{2^{37}}\)
MV
8 tháng 6 2017
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\\ \dfrac{1}{2\cdot2}\cdot\dfrac{2}{2\cdot3}\cdot\dfrac{3}{2\cdot4}\cdot\dfrac{4}{2\cdot5}\cdot...\cdot\dfrac{30}{2\cdot31}\cdot\dfrac{31}{2\cdot32}=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{30}{31}\cdot\dfrac{31}{32}\right)=2x\\ \dfrac{1}{2}\cdot\left(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot3\cdot4\cdot5\cdot...\cdot31\cdot32}\right)=2x\\ \dfrac{1}{2}\cdot\dfrac{1}{32}=2x\\ 2x=\dfrac{1}{64}\\ x=\dfrac{1}{64}:2\\ x=\dfrac{1}{128}\)
LT
1
HT
0
QM
1
2 tháng 5 2023
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
QM
1
12 tháng 5 2023
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
17 tháng 6 2022
a: =>2x-1=4 hoặc 2x-1=-4
=>2x=5 hoặc 2x=-3
=>x=5/2 hoặc x=-3/2
d: =>x=|2|=2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)
\(\left(\frac{-7}{2}\right)^2+\left(\frac{-3}{4}\right)^3.64-\left(\frac{-61}{5}\right)^0\)
\(=\frac{49}{4}+\frac{-27}{64}.64-1\)
\(=\frac{49}{4}-27-1\)
\(=\frac{-59}{4}-1\)
\(=\frac{-63}{4}\)