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\(3^{200}=9^{100}>4^{100}\\ 5^{200}=25^{100}< 64^{100}=4^{300}\\ 6^{50}=36^{25}>7^{25}\\ 8^{40}=64^{20}>10^{20}\\ 16^{20}=256^{10}>32^{10}\)
tick mik nha!!
3200=9100>41005200=25100<64100=4300650=3625>725840=6420>10201620=25610>3210
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
\(2,\)
\(a,20-\left[4^2+\left(x-6\right)\right]=90\)
\(\Rightarrow20-16-x+6=90\)
\(\Rightarrow10-x=90\)
\(\Rightarrow x=-80\)
Vậy: \(x=-80\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2^x-6\right)=25\)
\(\Rightarrow24+2^x=40\)
\(\Rightarrow2^x=16\)
\(\Rightarrow x=4\)
Vậy: \(x=4\)
\(2,\)
\(a,20-\left[42+\left(x-6\right)\right]=90\)
\(\Rightarrow20-42-x+6-90=0\)
\(\Rightarrow x=-106\)
Vậy: \(x=-106\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2x-6\right)=74\)
\(\Rightarrow1000:\left(24+2x\right)=74\)
\(\Rightarrow24+2x=\dfrac{500}{37}\)
\(\Rightarrow2x=-\dfrac{388}{37}\)
\(\Rightarrow x=-\dfrac{194}{37}\)
Mà \(x\in N\)
\(\Rightarrow x\in\varnothing\)
Vậy: \(x\in\varnothing\)
Ta có:
97 x 32 ⋮ 8
8 ⋮ 8
⇒ 97 x 32 + 8 ⋮ 8
Ta có:
2020 x 30 ⋮ 10
8 x 5 = 40 ⋮ 10
⇒ 2020 x 30 + 8 x 5 ⋮ 10
Bài 1:
a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)
b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)
c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)
d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)
e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)
Bài 2:
\(40-x=2^6\cdot2^2\)
\(\Rightarrow40-x=2^8\)
\(\Rightarrow40-x=256\)
\(\Rightarrow x=40-256\)
\(\Rightarrow x=-216\)
b) \(3^2\cdot3^x=81\)
\(\Rightarrow3^{2+x}=3^4\)
\(\Rightarrow2+x=4\)
\(\Rightarrow x=4-2=2\)
c) \(2^x=512\)
\(\Rightarrow2^x=2^9\)
\(\Rightarrow x=9\)
d) \(x^5=243\)
\(\Rightarrow x^5=3^5\)
\(\Rightarrow x=3\)
Bài 3:
a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)
b) \(8^3=\left(2^3\right)^3=2^9=512\)
c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)
d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)
e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)
\(=18-\left(4+3\right)\)
\(=18-7=11\)
2:
a: =>40-x=256
=>x=40-256=-216
b: =>x+2=4
=>x=2
c: =>2^x=2^9
=>x=9
d; =>x^5=3^5
=>x=3
\(42-\left(2x+32\right)-12:\left(-2\right)=6\)
\(42-2x-32+12:2=6\)
\(\left(42-32+6-6\right)-2x=0\)
\(10-2x=0\)
\(x=5\)
\(\left(x-15\right):\left(-5\right)-\left(-22\right)=24\)
\(\left(x-15\right):\left(-5\right)+22=24\)
\(\left(x-15\right):\left(-5\right)=2\)
\(x-15=2\cdot\left(-5\right)\)
\(x-5=-10\)
\(x=-5\)
42−(2x+32)−12:(−2)=6
42−2x−32+12:2=6
10−2x=0
(x−15):(−5)−(−22)=24
(x−15):(−5)+22=24
(x−15):(−5)=2
x−15=2·(−5)
x−5=−10
x=−5
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)