Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
Ta thấy \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3};......;\frac{1}{50^2}=\frac{1}{50.50}< \frac{1}{49.50}\)
Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=B\)
\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(B=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vì \(B< 2\)mà \(A< B\)nên \(A< 2\left(đpcm\right)\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}.\)Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\);...; \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
=> \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=> \(A< 1+1-\frac{1}{50}\)
=> \(A< 2-\frac{1}{50}\)
=> \(A< 2\)
\(4\left(3x^3+1^{10}\right)=4\cdot5^2\)
=>\(4\left(3x^3+1\right)=4\cdot25=100\)
=>\(3x^3+1=25\)
=>\(3x^3=24\)
=>\(x^3=8\)
=>\(x=\sqrt[3]{8}=2\)
\(4\cdot\left(3x^3+1^{10}\right)=4\cdot5^2\)
\(\Rightarrow3x^3+1=\dfrac{4\cdot5^2}{4}\)
\(\Rightarrow3x^3+1=5^2\)
\(\Rightarrow3x^3=25-1\)
\(\Rightarrow3x^3=24\)
\(\Rightarrow x^3=\dfrac{24}{3}\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
1)
a) -(2+5) = -2 - 5 = -7
b) +(-3+6) = -3 + 6 = 3
c) (-50+3) = -50 + 3 = -47
d) -(-2+3) = 2 - 3 = -1
e) -(10-3) = -10 + 3 = -7
f) -(-3)-(-3+1) = 3 + 3 - 1 = 5
g) (-5)+(-2+10) = -5 - 2 + 10 = 3
2)
a) -50+120+(-150)-20+30
= -(50 + 20) + (120 + 30 - 150)
= -70
b) 265-70+(-65)-30+15
= (265 - 65) - (70 + 30) + 15
= 200 - 100 + 15 = 115
c) -17+185-183+(-85)-63
= (185 - 85) - (183 + 17) - 63
= 100 - 200 - 63 = -163
d) -30+60+(-170)-260+19
= -(170 + 30) - (260 - 60) + 19
= -200 - 200 + 19 = -381
Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)
\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2017.2017}\)
Vì \(1=1\)
\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{2017.2017}< \frac{1}{2016.2017}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=2-\frac{1}{2017}< 2\)
\(\Rightarrowđpcm\)
(47 - 52). 3 + 27
= -5. 3 + 27
= -15 + 27
= 12
-4(39 - 19) + 30(-5 - 15)
= -4. 20 + 30. (-20)
= -80 - 600
= -680
36. (-9) + 9. (-64)
= -9(36 + 64)
= -9. 100
= -900
\(\dfrac{15}{17}+\dfrac{32}{17}=\dfrac{15+32}{17}=\dfrac{47}{17}\)
Câu 1: (5-25)-(-17)+40
=-20-(-17)+40
=-37+40
=3
Câu 2: (-7)-[(-30)+27]+(-11)
=(-7)-(-3)+(-11)
=-10+(-11)
=-21
Câu 3: (-77)+13+77+100+(-13)
=-64+77+100+(-13)
=13+100+(-13)
=113+(-13)
=100
Câu 4: 9-18+[15-(-28)]
=9-18+-13
=-9+-13
=-22
a) -17-(8-25)
= -17 - 8 + 25
= -25 + 25 =0
b)(-44)-(-14-30)
= -44 + 14 + 30
= -30 + 30 = 0
a) -17 - (8 - 25)
= -17 - 8 + 25
= -25 + 25
= 0
b) (- 44) - (-14 - 30)
= -44 + 14 + 30
= -30 + 30
= 0
30 - (5-1^1)
30 - (5-1)
30 - 4
26
giải hộ với
.