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a) \(11^9+12^9+13^9+14^9+15^9+16^9\)
\(=11^{4.2}.11+12^{4.2}.12+13^{4.2}.13+14^{4.2}.14+15^9+16^9\)
\(=...1.11+...6.12+...1.13+...6.14+...5+...6\)
\(=...1+...2+...3+...4+...5+...6\)
\(=...1\)
Vậy biểu thức trên có chũ số tận cùng là 1
b) \(25^7+26^7+27^7+28^7+29^7+29^7+30^7+31^7\)
\(=...5+...6+27^4.27^3+28^4.28^3+29^4.29^3+29^4.29^3+...0+...1\)
\(=...5+...6+...3+...8+...9+...9+...0+...1\)
\(=...1\)
Vậy biểu thức trên có chữ số tận cùng là 1
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) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
Xem cách làm câu (b);(c);(d)
Bạn tham khảo:
Câu hỏi của Nguyễn Ngọc Thảo My - Toán lớp 7 - Học toán với OnlineMath
các bạn giúp mik nha
Cho A bằng 5^2021+1 phần 5^2022+1 ; B bằng 5^2020+1 phần 5^2021+1. Hãy so sánh A và B
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!
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)