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a) (x^3+12):4=60+(-3)
(x^3+12).1/4=57
x^3+12=228
x^3=216
x^3=6^3
=> x=6
b) 2^x+1.3=96
2^x+1.3=2^5.3
2^x+1=2^5.3:3
2^x+1=2^5
=> x+1=5
x=4
( x + 34 ) . 215 = 212 . 64
( x + 81 ) . 215 = 212 . ( 23 )4
( x + 81) . 215 = 212 . 212
( x + 81 ) . 215 = 224
x + 81 = 224 : 215 = 29 = 512
x = 512 - 81 = 431
Bài 1:
a, 96 \(⋮x=>x\inư\left(96\right)\)
b, \(2^x.15+2^x.17=4^{30}\)
\(2^x\left(15+17\right)=4^{30}\)
\(2^x.32=4^{30}\)
\(2^x.2^5=2^{60}\)
\(2^x=2^{60}:2^5\)
\(2^x=2^{55}\)
=> x = 55
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
a)x-12=-28
x=(-28)+12
x=-16
vậy x=-16
b)20+8(x+3)=5^2.4
20+8(x+3)=25.4
20+8(x+3)=100
8(x+3)=100-20
8(x+3)=80
x+3=80:8
x+3=10
x=10-3
x=7
vậy x=7
\(a,x-12=-28\)
\(x=-28+12\)
\(x=-16\)
\(b,20+8\left(x+3\right)=5^2.4\)
\(20+8\left(x+3\right)=100\)
\(8\left(x+3\right)=100-20\)
\(8\left(x+3\right)=80\)
\(x+3=80:8\)
\(x+3=10\)
\(x=10-3\)
\(x=7\)
\(\Leftrightarrow2^x\cdot\dfrac{1}{8}+2^x\cdot\dfrac{1}{4}+2^x\cdot\dfrac{1}{2}=254\)
\(\Leftrightarrow2^x\cdot\dfrac{7}{8}=254\)
\(\Leftrightarrow2^x=\dfrac{2032}{7}\)
mà x là số tự nhiên
nên \(x\in\varnothing\)
a, => 3^x.(6+4.3) = 100-72
=> 3^x.18 = 18
=> 3^x = 18:18 = 1
=> 3^x = 3^0
=> x=0
b, => 5^x.(5+4.5) = -12+175-38
=> 5^x.25 = 125
=> 5^x = 125:25 = 5
=> 5^x = 5^1
=> x=1
Tk mk nha
a) \(6\cdot3^x+4\cdot3^{x+1}=100+\left(-72\right)\)
\(\Leftrightarrow6\cdot3^x+4\cdot3^x\cdot3=100-72\)
\(\Leftrightarrow6\cdot3^x+12\cdot3^x=28\)
\(\Leftrightarrow18\cdot3^x=28\Leftrightarrow3^x=\frac{14}{9}\)
Phần a) bn xem lại đề bài nhé!!
b) \(5\cdot5^x+4\cdot5^{x+1}=\left(-12\right)+175+\left(-38\right)\)
\(\Leftrightarrow5^{x+1}+4\cdot5^{x+1}=175-12-38\)
\(\Leftrightarrow5\cdot5^{x+1}=125\Leftrightarrow5^{x+1}=25\)
\(\Leftrightarrow5^{x+1}=5^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)
Vậy x=1
2^x.12=100
2^x.12=đề sai
đề đúng rùi pham minh quang