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a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)
a)x=2
b)x+8=14
x=6
c)\(42-x=5^2\)
\(42-x=25\)
\(-x=-17\)
\(x=17\)
+) \(x^3=x^2\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) \((7x-11)^3=2^5.5^2+200\)
\((7x-11)^3=2^3.2^2.5^2+2^3.5^2\)
\((7x-11)^3=2^3.5^2.(2^2+1)\)
\((7x-11)^3=2^3.5^2.5\)
\((7x-11)^3=2^3.5^3\)
\((7x-11)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(7x=21\)
\(x=3\)
+) \(3+2^{x-1}=24-[4^2-(2^2-1)]\)
\(3+2^{x-1}=11\)
\(2^{x-1}=8\)
\(2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(x=4\)
a) 22 ( x + 32 ) - 5 = 55
=>4(x+9)-5=55
=>4(x+9)=60
=>x+9=15
=>x=6
b)7 ( x + 52 ) - 20 = 190
=>7(x+25)=210
=>x+25=30
=>x=5
c)6 ( x + 23 ) + 40 = 100
=>6(x+8)+40=100
=>6(x+8)=60
=>x+8=10
=>x=2
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\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)