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1;Ta có\(5.3^x=5.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
2.Ta có \(9.5^x=6.5^6+3.5^6\)
\(\Rightarrow9.5^x=5^6.\left(6+3\right)\)
\(\Rightarrow9.5^x=9.5^6\)
\(\Rightarrow5^x=5^6\)\
\(\Rightarrow x=6\)
3, Ta có \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
\(\Rightarrow3^{x+1}.\left(2.3+4\right)=10.3^6\)
\(\Rightarrow3^{x+1}.10=10.3^6\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
a) 5.3x = 5.34
=> 3x=34
=> x=4
b) 9.5x=6.56+3.56
=> 9.5x = (6+3)56
=> 9.5x=9.56
=> 5x=56
=> x=6
c) 2.3x+2 + 4.3x+1 = 10.36
=> 2.3x+1.3 + 4.3x+1 = 10.36
=> 6.3x+1+4.3x+1=10.36
=> (6+4).3x+1=10.36
=> 10.3x+1=10.36
=> 3x+1=36
=> x+1=6
=> x=5
ta có
\(3^{1+2+3+..+x}=3^{3.12}\Leftrightarrow\frac{x\left(x+1\right)}{2}=36\)
\(\Leftrightarrow x.\left(x+1\right)=72=8.9\Leftrightarrow x=8\)
b. ta có
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}=\left(\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)+1-\frac{1}{5^{2017}}\)
\(=A+1-\frac{1}{5^{2017}}\Rightarrow4A=1-\frac{1}{5^{2017}}< 1\Rightarrow A< \frac{1}{4}\)
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
1: \(5\cdot3^x=5\cdot3^4\)
nên \(3^x=3^4\)
hay x=4
2: \(7\cdot4^x=7\cdot4^3\)
nên \(4^x=4^3\)
hay x=3
3: \(8\cdot7^x=8\cdot7^6\)
nên \(7^x=7^6\)
hay x=6
a, => 2^x = (2^3)^4/(2^4)^3 = 2^12/2^12 = 1 = 2^0
=> x = 0
c, => 4^x = 4^10.(4-3) = 4^10
=> x=10
d, => 2^2.3^x-1 + 2.3^x.9 = 2^2.3^6+2.3^9
=> 2.3^x-1 . (2+3.9) = 2.3^6.(2+3^3)
=> 2.3^x-1 . 27 = 2.3^6 . 27
=> 3^x-1 = 3^6
=> x-1 = 6
=> x = 7
e, => 2^x.(1/3+1/6+2) = 2^11.(2+1/2)
=> 2^x. 5/2 = 2^11. 5/2
=> 2^x = 2^11
=> x = 11
Tk mk nha
\(1.a)\dfrac{2^3+3.26-4^3}{2^3.3^2}\)
\(=\dfrac{2^3.3.2.13-\left(2^2\right)^3}{2^3.3^2}\)
\(=\dfrac{2^4.3.13-2^6}{2^3.3^2}\)
\(=\dfrac{2^3\left(2.3.13-2^3\right)}{2^3.3^2}\)
\(=\dfrac{78-8}{9}\)
\(=\dfrac{70}{9}\)
\(b)\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^4.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{13}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+2.5\right)}{2^{11}.3^{11}\left(2.3\right)}\)
\(=\dfrac{2.11}{3.6}\)
\(=\dfrac{11}{9}\)
\(2.3^{x-1}-3^{x+1}=90\)
\(\Leftrightarrow3^x:3-3^x.3=90\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}-3\right)=90\)
\(\Leftrightarrow3^x.\dfrac{-8}{3}=90\)
\(\Leftrightarrow3^x=\dfrac{-135}{4}\)
\(\Leftrightarrow\) \(x\) 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
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi