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1:
\(\Leftrightarrow4\cdot3^x\cdot\dfrac{1}{9}+2\cdot3^x\cdot3=4\cdot3^4+2\cdot3^7\)
\(\Leftrightarrow3^x\cdot\left(\dfrac{4}{9}+6\right)=3^4\cdot\left(4+2\cdot3^3\right)\)
\(\Leftrightarrow3^x=729\)
hay x=6
2: \(\Leftrightarrow3^x\cdot4\cdot\dfrac{1}{3}+3^x\cdot2\cdot9=4\cdot3^6+2\cdot3^9\)
\(\Leftrightarrow3^x\cdot\dfrac{58}{3}=42282\)
=>3x=2187
hay x=7
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A) \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
=> \(2.3.3^{x+1}+4.3^{x+1}=10.3^6\)
=> \(6.3^{x+1}+4.3^{x+1}=10.3^6\)
=> \(\left(6+4\right).3^{x+1}=10.3^6\)
=> \(10.3^{x+1}=10.3^6\)
=> \(3^{x+1}=3^6\)
=> \(x+1=6\)
=> \(x=6-1\)
=> \(x=5\)
Vậy \(x=5.\)
B) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
=> \(6.8^{x-1}+8^{x-1}.8^2=6.8^{19}+8^{19}.8^2\)
=> \(8^{x-1}.\left(6+8^2\right)=8^{19}.\left(6+8^2\right)\)
=> \(8^{x-1}=8^{19}\)
=> \(x-1=19\)
=> \(x=19+1\)
=> \(x=20\)
Vậy \(x=20.\)
Còn câu c) thì mình đang nghĩ nhé.
Chúc bạn học tốt!
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\(\left(x-7\right)^{x+1}-\left(x-7\right)^{16}=0\)
\(\Leftrightarrow\left(x-7\right)^{16}.\left(x-7\right)^{x-15}-\left(x-7\right)^{16}=0\)
\(\Leftrightarrow\left(x-7\right)^{16}\left[\left(x-7\right)^{x-15}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{16}=0\\\left(x-7\right)^{x-15}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x-15}=1^{x-15};\left(x-7\right)^{x-15}=\left(x-7\right)^0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x-7=1;x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=8;x=15\end{cases}}\)
Vậy \(x\in\left\{7;8;15\right\}\)
P/s: Thay cái ngoặc có 2 nhánh thành ngoặc 3 nhánh cho nó đẹp :))
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2.3x + 3x - 1 = 7 . (32 + 2 . 62)
=> 2.3x + 3x - 1 = 567
=> 7 . 3x - 1 = 567
=> 3x - 1 = 567 : 7 = 81
=> x - 1 = 4
=> x = 5
a)2*3x+3x-1=7(32+2*62)
2*3x+3x-1=7(9+72)=7*81
2*3x+3x/3=567
2*3x+3x*1/3=567
(2+1/3)*3x=567
7/3*3x=567
3x=567:7/3
3x=243=35
=>x=5
b) mk ko hiểu đề mấy, cái chỗ 7x+2 là nhân vs 2 ak
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f(x) có :
\(\frac{101-1}{2}+1=51\) (số hạng)
\(\Rightarrow f\left(1\right)=1+1^3+1^5+1^7+...+1^{101}\)
\(=1+1+1+1+...+1\)
\(=51\)
\(f=\left(-1\right)=1+\left(1\right)^3+\left(-1\right)^5+\left(-1\right)^7+...+\left(-1\right)^{101}\)
\(=1-1-1-1-...-1\)
\(=-49\)
~ Học tốt ~
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Bài1:
a)Ta có:
\(-203< 0;\dfrac{1}{2017}>0\)
Nên \(-203< \dfrac{1}{2017}\)
b)\(\dfrac{7}{29}và\dfrac{12}{47}\)
c)Đặt \(A=\dfrac{10^{11}+1}{10^{12}+1}\);\(B=\dfrac{10^{12}+1}{10^{13}+1}\)
Ta có:\(10A=\dfrac{10^{12}+1+9}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)
\(10B=\dfrac{10^{13}+1+9}{10^{13}+1}=1+\dfrac{9}{10^{13}+1}\)
Do đó:\(10A>10B\Rightarrow A>B\)
Bài2:
a)\(500>2^x>100\)
Ta có:\(100< 2^7< 2^8< 500\)
\(\Rightarrow x\in\left\{7;8\right\}\)
Vậy...
Câu sau tương tự
a) Ta có: \(-203< 0;\dfrac{1}{2017}>0\)
\(\Rightarrow\dfrac{1}{2017}>-203\)
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4) \(2.3^x+3^{x-1}=7.\left(3^2+2.6^2\right)\)
\(\Rightarrow2.3^x+3^{x-1}=567\)
\(\Rightarrow7.3^{x-1}=567\)
\(\Rightarrow3^{x-1}=567\div7\)
\(\Rightarrow3^{x-1}=81\)
\(\Rightarrow3^{x-1}=3^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=4+1\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
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=>(x-7)x+1=(x-7)x+11
=>x+1=x+11
=>0x=10
=>0=10(vô lí)
vậy ko có giá trị nào của x thỏa mãn đề bài
`2 . 3^(x - 1) - 7 = 11`
`=> 2. 3^(x - 1) = 11 + 7`
`=> 2 . 3^(x - 1) = 18`
`=> 3^(x - 1) = 18:2`
`=> 3^(x-1) = 9`
`=> 3^(x - 1) = 3^2`
`=> x-1=2`
`=> x=3`
Vậy: `x=3`