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Bài 1:
\(\left(x-2013\right)^{2014}=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2013=1\\x-2013=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2014\\x=2012\end{cases}}}\)
Vậy x=2014; x=2012
Bài 2:
a) Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Ta thấy 8<9 => \(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
b) Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Ta thấy \(3^{2009}< 3^{2010}\Rightarrow3^{2009}< 9^{1005}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Thấy \(9801< 9999\Rightarrow9801^{10}< 9999^{10}\Rightarrow99^2< 9999^{10}\)
B1: (x-2013)2014=1 =>x-2013=1;-1=>x=2014;2012 B2: a)có:2333=(23)111=8111 ; 3222=(32)111=9111 =>2333<3222(8111<9111) b)có:91005=(32)1005=32010 >32009 =>91005>32009 c)có:9920=(992)10=980110<999910 =>9920<999910
Bạn tham khảo nhé
a ) Ta có :
\(\left(-\frac{1}{5}\right)^{300}=\left(\frac{1}{5}\right)^{300}=\frac{1}{5^{300}}=\frac{1}{\left(5^3\right)^{100}}=\frac{1}{125^{100}}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(\frac{1}{3}\right)^{500}=\frac{1}{3^{500}}=\frac{1}{\left(3^5\right)^{100}}=\frac{1}{243^{100}}\)
Do \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\left(125^{100}< 243^{100}\right)\)
\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)
b )
Ta có :
\(2550^{10}=\left(50.51\right)^{10}=50^{10}.51^{10}\)
\(50^{20}=50^{10}.50^{10}\)
Do \(50^{10}.51^{10}>50^{10}.50^{10}\)
\(\Rightarrow50^{20}< 2550^{10}\)
c )
Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(5^{50}=\left(5^2\right)^{25}=25^{25}\)
Do \(16^{25}< 25^{25}< 27^{25}\)
\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)
đặt
\(\frac{x}{5}=\frac{y}{4}=k=>x=5k,y=4k\)
\(=>x^2.y=25k^2.4k=100\)
\(k^3=1=>k=1\)
\(=>x=5,y=4\)
Vậy x=5, y=4
Bài 1: HS tự làm
Bài 2:
\(ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow a=c\left(a,b,c\ne0\right)^{\left(1\right)}\)
\(ab=c^2\Rightarrow\frac{a}{c}=\frac{c}{b}\Rightarrow a=b\left(a,b,c\ne0\right)^{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow a=b=c\)
\(\Rightarrow\frac{b^{3333}}{a^{1111}c^{2222}}=\frac{b^{3333}}{a^{1111+2222}}=\frac{b^{3333}}{a^{3333}}=1\)
a, Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
b, Ta có : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
\(\Rightarrow3^{2009}< 9^{1005}\)
c, Ta có : \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
a) Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì 9>8 nên 9111>8111
Vậy 3222>2333
b) Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Vì 2010>2009 nên 32010>32009
Vậy 91005>32009
c)Ta có:\(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 nên (99.99)10<(99.101)10
Vậy 9920<999910
a) \(2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8< 9\)\(\Rightarrow8^{111}< 9^{111}\)\(\Rightarrow2^{333}< 3^{222}\)
b) \(9^{1005}=\left(3^2\right)^{1005}=3^{2.1005}=3^{2010}>3^{2009}\)
a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{34}{103}\)
b) \(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{2000.1999}-\left(\frac{1}{1999.1998}+\frac{1}{1998.1997}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)(*)
Đặt biểu thức trong ngoặc là A ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}\)
\(A=1-\frac{1}{1999}\)
\(A=\frac{1998}{1999}\)
Thay vào biểu thức (*) ta có :
\(\frac{1}{2000.1999}-\frac{1998}{1999}\)
\(=\frac{1}{3998000}-\frac{1998}{1999}\)
\(=\frac{-3995999}{3998000}\)
c) \(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(=\frac{-1}{1.3}+\frac{-1}{3.5}+\frac{-1}{5.7}+\frac{-1}{7.9}+...+\frac{-1}{99.101}\)
\(=\frac{-1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\frac{100}{101}\)
\(=\frac{-50}{101}\)
_Chúc bạn học tốt_
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
a) \(\frac{5}{9}:\left(\frac{5}{12}-\frac{1}{11}\right)-\frac{5}{9}:\left(\frac{-1}{5}-\frac{2}{3}\right)\)
= \(\frac{5}{9}:\left(\frac{55}{132}-\frac{12}{132}\right)-\frac{5}{9}:\left(\frac{-3}{15}-\frac{10}{15}\right)\)
= \(\frac{5}{9}:\frac{43}{132}-\frac{5}{9}:\frac{-13}{15}\)
= \(\frac{5}{9}\times\frac{132}{43}-\frac{5}{9}\times\frac{-15}{13}\)
=\(\frac{5}{9}\times\left(\frac{132}{43}-\frac{-15}{13}\right)\)
=\(\frac{5}{9}\times\frac{2361}{559}\)( Đến đây bạn tự quy đồng mẫu nha)
=\(\frac{3935}{1677}\)
Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Do : \(8^{111}< 9^{111}\left(8< 9\right)\)
\(\Rightarrow2^{333}< 3^{222}\)
Ta có : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Do : \(3^{2009}< 3^{2010}\left(2009< 2010\right)\)
\(\Rightarrow3^{2009}< 9^{1005}\)