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a) $125^3:25^4=(5^3)^3:(5^2)^4=5^9:5^8=5^1$
$16^4:4^2=(4^2)^4:4^2=4^8:4^2=4^6$
$27^8:9^4=(3^3)^8:(3^2)^4=3^{24}:3^8=3^{16}$
$125^5:25^3=(5^3)^5:(5^2)^3=5^{15}:5^6=5^9$
$4^{14}:5^{28}=(2^2)^{14}:5^{28}=2^{28}:5^{28}=(\dfrac{2}{5})^{28}$
b) $12^n:2^{2n}=12^n:(2^2)^n=12^n:4^n=3^n$
$64^4.16^5.4^{20}=(4^3)^4.(4^2)^5.4^{20}=4^{12}.4^{10}.4^{20}=4^{42}$
1.
a. Ta có: \(A=2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(B=3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow A< B\)
b. Ta có: \(A=2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(B=3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Mà \(8^{111}< 9^{111}\)
\(\Rightarrow A< B\)
c. Ta có: \(A=2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(B=5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Mà \(8192^7>3125^7\)
\(\Rightarrow A>B\)
Câu 2:
a: =>(x-6)(x-7)=0
=>x=6 hoặc x=7
b: =>\(x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
a125^5:25^3=(5^3)^5:(5^2)^3=5^15:5^6=5^9
b27^6:9^3=(3^3)^6:(3^2)^3=3^18:3^6=3^13
c 4^20:2^15=(2^2)^20:2^15=2^40:2^15=2^25
d24^n:2^2.n=24^n:(2^2)^n=24^n:4^n=(24:4)^n=6^n
e 64^4 . 16^5:4^20=(2^6)^4 . (2^4)^5 :(2^2)^20=2^24 . 2^20:2^40=2^4
g 32^4:8^6=(2^5)^4:(2^3)^6=2^20:2^18=2^2
a, \(125^5:25^3=\left(5^3\right)^5:\left(5^2\right)^3=5^{15}:5^6=5^9\)
b, \(27^6:9^3=\left(3^3\right)^6:\left(3^2\right)^3=3^{18}:3^6=3^{12}\)
c, \(4^{20}:2^{15}=\left(2^2\right)^{20}:2^{15}=2^{40}:2^{15}=2^{25}\)
d, \(24^n:2^{2.n}=2^n.12^n:2^n.2^n=12^n:2^n=2^n.6^n:2^n=6^n\)
e, \(64^4.16^5:4^{20}=4^{12}.4^{10}:4^{20}=4^{12+10-20}=4^2\)
g, \(32^4:8^6=8^4.4^4:8^4.8^2=4^4:4^2.2^2=4^2.2^2=2^4.2^2=2^6\)
\(M=\dfrac{8^{20}+4^{20}}{4^{25}+64^5}\)
\(\Leftrightarrow M=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)
\(\Leftrightarrow M=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}\)
\(\Leftrightarrow M=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)
\(\Leftrightarrow M=\dfrac{2^{40}}{2^{30}}\)
\(\Leftrightarrow M=2^{10}.\)