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vậy mà cũng trả lời sao trời

A)\(5^3và3^5\)

\(5^3=125;3^5=243\)

Vì\(125< 243\Rightarrow5^3< 3^5\)

B)\(4^3va3^4\)

\(4^3=64;3^4=81\)

Vì\(64< 81\Rightarrow4^3< 3^4\)

C)\(2^4va8^2\)

\(2^4=\left(2^2\right)^2=4^2\)

Vì\(4^2< 8^2\Rightarrow2^4< 8^2\)

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

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

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

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

2:

a: A=1+2+2^2+2^3+2^4

=>2A=2+2^2+2^3+2^4+2^5

=>A=2^5-1

=>A=B

b: C=3+3^2+...+3^100

=>3C=3^2+3^3+...+3^101

=>2C=3^101-3

=>\(C=\dfrac{3^{101}-3}{2}\)

=>C=D

Ta có: 

\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)

\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)

A=\(\frac{5^5}{5+5^2+5^3+5^4}=\frac{5^5}{5\left(1+5+5^2+5^3\right)}=\frac{5^4}{1+5+25+125}\)=\(\frac{5^4}{1+155}=\frac{625}{156}\)

B=\(\frac{3^5}{3+3^2+3^3+3^4}=\frac{3^5}{3\left(1+3+3^2+3^3\right)}=\frac{3^4}{1+3+9+27}\)=\(\frac{3^4}{1+39}=\frac{81}{40}\)

Ta có:\(\frac{625}{156}\)>\(\frac{81}{40}\)\(\Rightarrow A\)>\(B\)

a)A=3^0+3^1+3^2+3^3+...+3^2012

A=1+3+3^2+3^3+..+3^2012

3A=3+3^2+3^3+3^4+..+3^2013

3A-A=3+3^2+3^3+3^4+..+3^2013-1-3-3^2-3^3-...-3^2012

2A=3^2013-1

A=\(\frac{3^{2013}-1}{2}\)

B=3^2013

=> A>B

b) A=1+5+5^2+5^3+..+5^99+5^100

5A=5+5^2+5^3+5^4+...+5^100+5^101

5A-A=5+5^2+5^3+5^4+..+5^100+5^101-1-5-5^2-5^3-..-5^99-5^100

4A=5^101-1

A=\(\frac{5^{101}-1}{4}\)

B=5^101/4

=> A<B

nhân 3A lên

nhân 5B lên