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3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
45^10*5^20/75^15
=5^10*9^10*5^20/(5^2)^15
=5^10*5^20*9^10/5^30
=9^10
(0.8)^5/(0.4)^6
=(0.4)^5*2^5/(0.4)^6
=2^5/(0.4)
=32/(0.4)
=80
2^15*9^4/6^6*8^3
=2^15*(3^2)^4/2^6*3^6*(2^3)^3
=2^15*3^8/2^6*3^6*2^9
=3^2
=9
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)