1
=>A=[tex]\left ( x+x+x+x+x \right )+\left ( \frac{1}{5}+\frac{4}{5} \right )+\left ( \frac{2}{5}+\frac{3}{5} \right )[/tex]
=[tex]5x+2[/tex]
vì 5x+2 > 5x
=>A > B
2
[tex]\frac{100}{3}+\frac{100}{3^{2}}+\frac{100}{3^{3}}+\frac{100}{3^{4}}[/tex]
[tex]=100.\left ( \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}} \right )[/tex]
[tex]=100.\left ( \frac{3^{3}}{3^{4}}+\frac{3^{2}}{3^{4}}+\frac{3}{3^{4}}+\frac{1}{3^{4}} \right )[/tex]
[tex]=100.\frac{40}{81}[/tex]
bài 3 làm tương tự bài 2 nha em
$A=[x]+[x+\dfrac{1}{5}]+[x+\dfrac{2}{5}]+[x+\dfrac{3}{5}]+[x+\dfrac{4}{5}]$ mà có phải bằng $x+(x+\dfrac{1}{5})+(x+\dfrac{2}{5})+(x+\dfrac{3}{5})+(x+\dfrac{4}{5})$ đâu nguyên ^^.
Bài 2 cx vậy ^^
1) Cho x =3,7 , so sánh A = [x] + [x+1/5] + [x+2/5] + [x+3/5] + [x+4/5] và B = [5x]
2) Tính: [100/3] + [100/(3^2)] + [100/(3^3)] +[100/(3^4)]
3) Tính :[50/2] + [50/(2^2)] + [50/(2^3)] + [50/(2^4)] + [50/(2^5)]
P/s:Hơi khó hỉu bởi vì ko bít vít p/số, mong m.n thông cảm
$1)A=[3,7]+[3,7+\dfrac{1}{5}]+[3,7+\dfrac{2}{5}]+[3,7+\dfrac{3}{5}]+[3,7+\dfrac{4}{5}]
\\=[3,7]+[3,9]+[4,1]+[4,3]+[4,5]=3+3+4+4+4=18\\
B=[5.3,7]=[18,5]=18
\\\Rightarrow A=B$
$2)[\dfrac{100}{3}] + [\dfrac{100}{3^2}] + [\dfrac{100}{3^3}] +[\dfrac{100}{3^4}]
\\=[\dfrac{99+1}{3}]+[\dfrac{99+1}{3^2}]+[\dfrac{81+19}{3^3}]+[\dfrac{81+19}{3^4}]
\\=[33+\dfrac{1}{3}]+[11+\dfrac{1}{3^2}]+[3+\dfrac{19}{3^3}]+[1+\dfrac{19}{3^4}]
\\=33+11+3+1=48$
$3)[\dfrac{50}{2}] + [\dfrac{50}{2^2}] + [\dfrac{50}{2^3}] + [\dfrac{50}{2^4}] + [\dfrac{50}{2^5}]
\\=[25]+[\dfrac{48+2}{2^2}]+[\dfrac{48+2}{2^3}]+[\dfrac{48+2}{2^4}]+[\dfrac{32+18}{2^5}]
\\=[25]+[12+\dfrac{2}{2^2}]+[6+\dfrac{2}{2^3}]+[3+\dfrac{2}{2^4}]+[1+\dfrac{18}{2^5}]
\\=25+12+6+3+1=47$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
