\(2^4.32>2^x>64\)
tìm x biết
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a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
\(2.x.2.2.2+2.2.2.x=64\)
\(2^4x+2^3x=64\)
\(x\left(2^4+2^3\right)=64\)
\(x\left(16+8\right)=64\)
\(24x=64\)
\(x=\frac{8}{3}\)
vay \(x=\frac{8}{3}\)
\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)
\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)
\(\Leftrightarrow12+9+112=64\)(vô lí)
Vậy pt vô nghiệm
TL:
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)
\(\Leftrightarrow6x+133=64\)
\(\Leftrightarrow6x=-69\)
\(\Leftrightarrow x=\frac{-23}{2}\)
Vậy....
1) (-10)-|5-x|=(-12)
|5-x|=(-10)-(-12)
|5-x|=2
\(\Rightarrow\left[{}\begin{matrix}5-x=2\\5-x=\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy:............
\(\Rightarrow\)\(2^{3+x}\)\(=\)\(2^{7776}\)
\(\Rightarrow\)\(3+x\)\(=7776\)
\(\Rightarrow\)\(x=7773\)
Vậy x = 7773
64^5=1073741824
2^3=8
1073741824:8=134217728
=>x=27 vì 227=134217728
nhớ k cho mình nha
a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
\(2^4.32>2^x>64\\ \Leftrightarrow2^4.2^5>2^x>2^6\\ \Leftrightarrow2^9>2^x>2^6\\ \Leftrightarrow9>x>6\)
=> x=7 hoặc x=8
2⁴.32>2^x>64
=>2⁹>2^x>2⁶
=>9>x>6
=>x€{8;7}