Ta có: `8^111 =(2^3 )^111 =2^(3.111)=2^333`

`4^170 =(2^2 )^170 =2^(2.170)=2^340`

Vì `333<340=>8^111 <4^170`

Ta có: `3^300 =3^(3.100)=(3^3 )^100=27^100`

`5^200 =5^(2.100)=(5^2 )^100 =25^100`

Vì `27>25=>3^300 >5^200`

a: 8^111=2^333

4^170=(2^2)^170=2^340

mà 333<340

nên 8^111<4^170

b: 3^300=(3^3)^100=27^100

(5^200)=(5^2)^100=25^100

mà 27>25

nên 3^300>5^200

a,Tính tổng:S=1+52+54+...+5200

=>5²S=5²+5⁴+5⁶+...+5²⁰²

=>25S-S=24S=5²⁰²-1

=>S=\(\frac{5^{202}-1}{24}\)

b,So sánh 230+330+430 và 3.2410

3.24^10=3^11.4^15 
4^30=4^15.4^15 
hiển nhiên 4^15>3^11 
=>3.24^10<<4^30<<<2^30+3^20+4^30

Ta có: 2³⁰+3³⁰+4³⁰>2³⁰+2³⁰+4³⁰=2³¹+2³⁰.2³⁰

                                                                 =2³¹(1+2²⁹) (1)

Lại có:3.24^10=3^11.2^30 (2)

So sánh (1)và (2): Vì 3^11<4^11=2^22<2^29

                              và 2^30<2^31

=> 3^11.2^30 <(1+2^29)2^31<2^30+3^30+4^30

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

42:x=6

x= 42 :6 

X= 7

TH 2

36:x = 6

X = 36: 6

X= 6

bài này dư thì phải

a) ta có:  \(1-\frac{2012}{2013}=\frac{1}{2013}\)

                 \(1-\frac{2013}{2014}=\frac{1}{2014}\)

mà \(\frac{1}{2013}>\frac{1}{2014}\) nên   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta

a/ ta có : a<b

=> 2a<2b

=>2a-1<2b-1

a) Hai cạnh bên BC = AD (=3 cm).

b) AB song song với CD.

c) AC = BD (=4,8 cm).

\(\frac{22}{27}+\frac{37}{67}=\left(1-\frac{5}{27}\right)+\left(1-\frac{30}{67}\right)\)

\(\frac{31}{36}+\frac{377}{677}=\left(1-\frac{5}{36}\right)+\left(1-\frac{300}{677}\right)\)

+ \(\frac{5}{27}>\frac{5}{36}\Rightarrow1-\frac{5}{27}< 1-\frac{5}{36}\left(1\right)\)

+ \(\frac{30}{67}=\frac{300}{670}>\frac{300}{677}\Rightarrow1-\frac{30}{67}< 1-\frac{300}{677}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{22}{27}+\frac{37}{67}< \frac{31}{36}+\frac{377}{677}\)

Vsu