\(x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{4}{5}=5x+\frac{10}{5}\)

\(5x+2>5x\)

\(A>B\)

\(x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{5}{5}=5x+\frac{10}{5}\)

\(5x+2>5x\)

\(A>B\)

\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

\(A=\left(x\right)+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)

\(A=\left(x+x+x+x+x\right)+\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)\)

\(A=5x+2>5x\Rightarrow A>B\)

\(A=x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{4}{5}\) 

\(=x+x+x+x+x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\) 

\(=5x+\frac{10}{5}\) 

\(=5x+2\) 

Vì 5x + 2 > 5x 

Vậy A > B 

\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)

\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)

\(=5x+2\)

\(B=5x\)

\(\Rightarrow A>B\)Với \(\forall\)\(x\)

#)Giải :

\(A=\left[x\right]+\left[1+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)

Thay x = 3,7 vào biểu thức, ta có :

\(A=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)

\(A=\left[3,7+3,7+3,7+3,7+3,7\right]+\left[1+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right]\)

\(A=18,5+3\)

\(A=21,5\)

\(B=\left[5x\right]=\left[5\times3,7\right]=18,5\)

Vì 21,5 > 18,5 \(\Rightarrow A>B\)

Bằng nhau.

A=5x+2;B=5x

mà x >0 do đó A>B

Giải

xét vế A :

thay x=3,7 vào biểu thức ta có:

A=[ 3,7   ]+[  3,7+1/5]+[3,7+2/5]+[3,7+3/5]+[ 3,7+4/5]

=(3.7*5)+(1/5+2/5+3/5+4/5)=20,5

xét vế B

thay x=3,7 vào biểu thức ta có

B=[5x]

=>b=[5*3.7]=5.3,7=18,5

+, ta có A=20,5 ; B=18,5

=>A>B

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)