a) \(A=\sqrt{140+1}và\sqrt{140}+\sqrt{1}\)

Ta có: \(\sqrt{140+1}=\sqrt{141}\approx11.87\)

\(\sqrt{140}+\sqrt{1}\approx12.83\) 11.8

\(\Rightarrow\) \(11.87>12.83\)

\(\Rightarrow\) \(\sqrt{140+1}>\sqrt{140}+1\)

b) \(A=\sqrt{222+2}\) và \(B=\sqrt{222}+\sqrt{2}\)

Ta có : \(A=\sqrt{222}+2\)

\(=\sqrt{224}\approx15\)

\(B=\sqrt{222}+\sqrt{2}\)

\(B=\sqrt{222}+\sqrt{2}\approx16.31\)

\(\Rightarrow\) \(A< B\)

NOTE: Đây cũng là cách giải của lớp 9 nên e xem thử để hỏi a nha

1)Ta có:\(ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c},ab=c^2\Rightarrow\frac{c}{a}=\frac{b}{c}\)

\(\Rightarrow\frac{a}{b}=\frac{c}{a}=\frac{b}{c}=\frac{a+c+b}{b+a+c}=1\)(T/C...)

\(\Rightarrow a=b=c\)

\(\Rightarrow M=\frac{b^{333}}{a^{111}\cdot c^{222}}=\frac{b^{333}}{b^{111}\cdot b^{222}}=\frac{b^{333}}{b^{333}}=1\)

chịu r

\(B=-1+\frac12-\frac{1}{2^2}+\cdots-\frac{1}{2^{2n}}+\frac{1}{2^{2n+1}}\)

=>\(2B=-2+1-\frac12+\cdots-\frac{1}{2^{2n-1}}+\frac{1}{2^{2n}}\)

=>\(2B+B=-2+1-\frac12+\cdots-\frac{1}{2^{2n-1}}+\frac{1}{2^{2n}}-1+\frac12-\frac{1}{2^2}+\cdots-\frac{1}{2^{2n}}+\frac{1}{2^{2n+1}}\)

=>\(3B=-2+\frac{1}{2^{2n+1}}=\frac{-2^{2n+2}+1}{2^{2n+1}}\)

=>\(B=\frac{-2^{2n+2}+1}{3\cdot2^{2n+1}}\)

a: \(\left(-1\right)\left(-1\right)^2\cdot\left(-1\right)^3\cdot\ldots\cdot\left(-1\right)^{2014}\)

\(=\left(-1\right)^{1+2+\cdots+2014}\)

\(=\left(-1\right)^{2014\cdot\frac{2015}{2}}=\left(-1\right)^{1007\cdot2015}=-1\)

b: \(\frac17\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{130}+\frac{13}{60}\right)\)

\(=\frac17\left(\frac52+\frac{4}{11}+\frac{3}{22}+\frac{11}{130}+\frac{13}{60}\right)\)

\(=\frac17\left(3+\frac{13}{60}+\frac{11}{130}\right)=\frac17\cdot\frac{515}{156}=\frac{515}{1092}\)

1. Tính 

\(\frac{2^7\cdot9^3}{6^5\cdot8^2}=\frac{2^7\cdot\left(3^2\right)^3}{\left(3\cdot2\right)^5\cdot\left(2^3\right)^2}=\frac{2^7\cdot3^6}{3^5\cdot2^5\cdot2^6}=\frac{2^7\cdot3^5\cdot3}{3^5\cdot2^{11}}=\frac{2^7\cdot3}{2^7\cdot2^4}=\frac{3}{2^4}=\frac{3}{16}\)

2. Tìm x

\(8^x:4^x=4\Rightarrow\left(8:4\right)^x=4\Rightarrow2^x=4\Rightarrow x=2\)

3. Ta có : \(222^{555}=\left(2\cdot111\right)^{555}=2^{555}\cdot111^{555}=\left(2^5\right)^{111}\cdot111^{555}=32^{111}\cdot111^{555}\)(1)

\(555^{222}=\left(5\cdot111\right)^{222}=5^{222}\cdot111^{222}=\left(5^2\right)^{111}\cdot111^{222}=25^{111}\cdot111^{222}\)(2)

Từ (1) và (2) ta thấy : 32 > 25 => 32¹¹¹ > 25¹¹¹ và 111⁵⁵⁵ > 111²²² ( vì 555 > 222)

Vậy 222⁵⁵⁵ > 555²²²

giúp mình nha 

-Vì (1/222)^333=(1/222)^3.111=(3/666)^111

     (1/333)^222=(1/333)^2.111=(2/666)^111

-Vì 111=111 và 3/666>2/666

=))(1/222)^333>(1/333)^222

bài 1 :
b) (x-1/2 )² = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) ² = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )³ = -8
<=> ( 2x - 1) ³ = ( -2 ) ³
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2

Bài 2 :
c) 3^2x-1=243
<=> 3^2x-1= 3⁵
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3

Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !

mấy câu kia cần nữa k

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....