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

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

\(=5x+2\)

\(B=5x\)

\(\Rightarrow A>B\)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 \(\Rightarrow A>B\)

Phạm Thị Thùy Linh+๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ ( Team TST 14 ):Cả 2 bạn đều nhầm chỗ  \(\left[a\right]\) rồi nha.\(\left[a\right]\) tức là phần nguyên của a nghĩa là số nguyên lớn nhất ko vượt quá a.

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

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

\(=3+3+4+4+4\)

\(=18\)

\(B=\left[5x\right]\)

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

\(=18\)

Vậy \(A=B\left(=18\right)\)

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

Bằng nhau.

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

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

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

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

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

\(=5x+\frac{10}{5}\) 

\(=5x+2\) 

Vì 5x + 2 > 5x 

Vậy A > B 

Mình cảm ơn bạn Capheny bản quyền nha

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

Thay \(3,7=3\frac{7}{10}\)vào biểu thức:

A = \(\left[3+\frac{7}{10}\right]+\left[3+\frac{9}{10}\right]+\left[3+\frac{11}{10}\right]+\left[3+\frac{13}{10}\right]+\left[3+\frac{15}{10}\right]\)

A = 3 + 3 + 4 +4 + 4 = 18

B = \(\left[5x\right]=\left[5.3,7\right]=\left[18,5\right]=18\)

Vậy A = B

1) c)

\(\left[\frac{1000}{3}\right]+\left[\frac{1000}{3^2}\right]+\left[\frac{1000}{3^3}\right]+\left[\frac{1000}{3^4}\right]=33+11+3+1=48\)