\(3^4\cdot3^x+3^{x+1}=243\)
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3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
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3^x=3^(1+2+3+....+100)
x=1+2+3+..+100
x=(100+1).100/2=5050
\(3^x=3^1\cdot3^2\cdot...\cdot3^{100}\)
\(3^x=3^{1+2+...+100}\)
\(3^x=3^{5050}\)
Vậy x = 5050
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\(1.a)\dfrac{2^3+3.26-4^3}{2^3.3^2}\)
\(=\dfrac{2^3.3.2.13-\left(2^2\right)^3}{2^3.3^2}\)
\(=\dfrac{2^4.3.13-2^6}{2^3.3^2}\)
\(=\dfrac{2^3\left(2.3.13-2^3\right)}{2^3.3^2}\)
\(=\dfrac{78-8}{9}\)
\(=\dfrac{70}{9}\)
\(b)\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^4.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{13}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+2.5\right)}{2^{11}.3^{11}\left(2.3\right)}\)
\(=\dfrac{2.11}{3.6}\)
\(=\dfrac{11}{9}\)
\(2.3^{x-1}-3^{x+1}=90\)
\(\Leftrightarrow3^x:3-3^x.3=90\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}-3\right)=90\)
\(\Leftrightarrow3^x.\dfrac{-8}{3}=90\)
\(\Leftrightarrow3^x=\dfrac{-135}{4}\)
\(\Leftrightarrow\) \(x\) không có giá trị nào để thỏa mãn đề bài.
Vậy \(x\in\varnothing\)
nữ thám tử nổi tiếng
Đề bài câu 2 sai thì phải, nếu đề bài đc sửa lại là \(3^{x-1}+3^{x+1}=90\) thì \(x=3\) có lẽ là đúng
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a) 15 - 3x = 6
=> 3x = 15 - 6
=> 3x = 9
=> x = 9 : 3 = 3
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{2}{3}-\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{6}:\dfrac{4}{3}\Rightarrow x=\dfrac{3}{24}=\dfrac{1}{8}\)c) 319.(x - 12 ) = 4 . 320
=> x - 12 = 4 . 3
=> x - 12 = 12
=> x = 12 + 12 = 24
d) 3.(x - 4) = 2^2 . 3^3
=> x - 4 = 2^2 . 3^2
=> x - 4 = 36
=> x = 36 + 4 = 40
a)\(15-3x=6\Rightarrow3x=9\Rightarrow x=3\)
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{8}\)
c) \(3^{19}.\left(x-12\right)=4.3^{20}\Rightarrow x-12=12\Rightarrow x=24\)
d)\(3\left(x-4\right)=2^2.3^3\Rightarrow3x-12=108\Rightarrow3x=120\Rightarrow x=40\)
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\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
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a) \(3^{2x-1}=243\)
\(\Leftrightarrow3^{2x-1}=3^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)
\(\Leftrightarrow3^{2x}:3^3=\dfrac{1}{3^5}\)
\(\Leftrightarrow3^{2x}:3^3=3^{-5}\)
\(\Leftrightarrow3^{2x-3}=3^{-5}\)
\(\Leftrightarrow2x-3=-5\)
\(\Leftrightarrow2x=-5+3\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{2}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
c) \(2^{3x+2}=4^{x+5}\)
\(\Leftrightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\)
\(\Leftrightarrow3x+2=2\left(x+5\right)\)
\(\Leftrightarrow3x+2=2x+10\)
\(\Leftrightarrow3x-2x=10-2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
d) \(3^{x+1}=9^x\)
\(\Leftrightarrow3^{x+1}=\left(3^2\right)^x\)
\(\Leftrightarrow3^{x+1}=3^{2x}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow2x-x=1\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
34 . 3x + 3x + 1 = 243
34 . 3x + 3x . 31 = 243
3x . ( 34 + 31 ) = 243
3x . ( 81 + 3 ) = 243
3x . 84 = 243
3x = 243 : 84
3x = .............. = 3....
Suy ra x = .......
Good luck. ( Phần cuối tự tính )