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\(A=3+3^2+....+3^{99}\)
\(3A=3^2+3^3+...+3^{100}\)
\(3A-A=3^2+3^3+...+3^{100}-3-3^2-...-3^{99}\)
\(2A=3^{100}-3\)
\(A=\dfrac{3^{100}-3}{2}\)
\(\Rightarrow2A+3=9^{2x+6}\)
\(\Rightarrow2\cdot\dfrac{3^{100}-3}{2}+3=\left(3^2\right)^{2x+6}\)
\(\Rightarrow3^{100}-3+3=3^{2\left(2x+6\right)}\)
\(\Rightarrow3^{100}=3^{4x+12}\)
\(\Rightarrow4x+12=100\)
\(\Rightarrow4x=88\)
\(\Rightarrow x=22\)
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Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
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Bài làm
\(3\cdot5^x-17=7^6\div7^3+315\)
\(3\cdot5^x-17=7^3+315\)
\(3\cdot5^x-17=343+315\)
\(3\cdot5^x-17=658\)
\(3\cdot5^x=658+17\)
\(3\cdot5^x=675\)
\(5^x=675:3\)
\(5^x=225\)
Hmmm... Đề có bị sai không ạ? Vì 53 = 125; 54 = 625. Nhưng kết quả của 5x = 225. Nên bị lỗi :/
Kiểm tra lại đề xong gửi đề cho mình để mình làm lại cho
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\(3^x+4^2=19^6:\left(19^3.19^2\right)-2.1^{2014}\)
\(\Rightarrow\) \(3^x+16=19^6:19^5-2\)
\(\Rightarrow\) \(3^x+16=19-2\)
\(\Rightarrow\) \(3^x+16=17\)
\(\Rightarrow\) \(3^x=1\)
\(\Rightarrow\) \(3^x=3^0\)
\(\Rightarrow\) \(x=0\)
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2x + 2x + 2 = 82 - 62 + 22 . 3
= 2x + 2x + 2 = 64 - 36 + 4 . 3
= 2x + 2x + 2 = 64 - 36 + 12
= 2x + 2x + 2 = 28 + 12
= 2x + 2x + 2 = 40
= 2x + 2x = 40 - 2
= 2x + 2x = 38
= 2x . 2 = 38
= 2x = 38 : 2
= 2x = 19
=> x không có giá trị
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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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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
\(\left(x-6\right)^3=\left(x-6\right)^2\)
Đặt \(x-6=a\)\(\Leftrightarrow a^3=a^2\)\(\Leftrightarrow a^3-a^2=0\)\(\Leftrightarrow a^2\left(a-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a^2=0\\a-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-6=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)(nhận)
Vậy \(x\in\left\{6;7\right\}\)