Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
a) \(3.\left(10.x\right)=111\)
\(10.x=37\)
\(x=\dfrac{37}{10}\)
b) \(3.\left(10+x\right)=111\)
\(10+x=37\)
\(x=27\)
c) \(3+\left(10.x\right)=111\)
\(10.x=108\)
\(x=\dfrac{54}{5}\)
d) \(3+\left(10+x\right)=111\)
\(x=111-3-10\)
\(x=98\)
\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)
\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)
\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)
\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)
\(=\dfrac{7}{2\left(1+6\right)}\)
\(=\dfrac{7}{2.7}\)
\(=\dfrac{1}{2}\)
a) \(5^{20}và2550^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)
=> \(5^{20}< 2550^{10}\)
b) \(999^{10}và999999^5\)
\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)
=> \(999^{10}< 999999^5\)
c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)
\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)
\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)
\(\dfrac{-1}{5}=\dfrac{-3}{15}\)
\(\dfrac{-1}{3}=\dfrac{-5}{15}\)
=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)
=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)
Bài 2:
a)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=> a = b = c
b)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
=> x = y = z (theo a)
Thay x = y = z vào biểu thức, ta có:
\(M=\dfrac{x^{333}.x^{666}}{x^{999}}=1\)
c)
\(ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(ab=c^2\Rightarrow\dfrac{b}{c}=\dfrac{c}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Rightarrow a=b=c\)
Thay a = b = c vào biểu thức, ta có:
\(M=\dfrac{a^{333}}{a^{111}.a^{222}}=1\)
\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\\ \left(\dfrac{1}{2}\right)^{300}=\left(\dfrac{1}{2}\right)^{3\cdot100}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\\ \left(\dfrac{1}{3}\right)^{200}=\left(\dfrac{1}{3}\right)^{2\cdot100}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\\ \dfrac{1}{8}>\dfrac{1}{9}\Rightarrow\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\Rightarrow\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\\ \left(0,3\right)^{20}=\left(0,3\right)^{2\cdot10}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}< \left(0,1\right)^{10}\)
a) \(\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\)
Vì \(40< 50\)
b)\(\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
\(\Rightarrow\text{}\text{}\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
Vì \(\dfrac{1}{8}>\dfrac{1}{9}\)
c)\(\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)
\(\Rightarrow\left(0,1\right)^{10}>\left(0,3\right)^{20}\)
Vì \(0,1>0,09\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Do : \(8^{111}< 9^{111}\left(8< 9\right)\)
\(\Rightarrow2^{333}< 3^{222}\)
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
ỦA SAO BẠN GIẢI RA LUÔN RỒI. HAY Zậy?
đúng là hay quá ha.
tại mình ấn nhầm