Ta thấy : \(\frac{1}{11}>\frac{1}{100},\frac{1}{12}>\frac{1}{100},...,\frac{1}{100}=\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}=1\)

Do đó : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>1\)

1999/2001 < 12/11

vì 1999/2001 bé hơn 1 còn 12/11 thì lớn hơn 1

1/a-1 < 1/a+1 ( bạn cho ví dụ thì dễ hơn ạ )

Ta có:

 \(\dfrac{1}{5}>\dfrac{1}{10}\\ \dfrac{1}{6}>\dfrac{1}{10}\\ ...\\ \dfrac{1}{9}>\dfrac{1}{10}\\ \Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{5}{10}=\dfrac{1}{2}.\)

Tương tự:

 \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}>\dfrac{5}{15}=\dfrac{1}{3}.\\ \dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}>\dfrac{3}{18}=\dfrac{1}{6}.\)

Cộng vế theo vế ta được \(B>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\left(đpcm\right)\)

a, -(-5)-(+7)+(+3)+(-8)

= 5 - 7 + 3 + (-8)

= (-2) + 3 + (-8)

= 1 + (-8)

= -7

b, -(-15)-|-10|+|-9|-|5I

= 15 - 10 + 9 - 5

= 5 + 9 - 5

= 14 - 5

= 9

c, 14-(-13)-(17)+(-12)

= 14 + 13 - 17 + (-12)

= 27 - 17 + (-12)

= 10 + (-12)

= 2

d, -|-14| + |-10| - (-12) + (-8)

= -14 + 10 + 12 + (-8)

= (-4) + 12 + (-8)

= 7 + (-8)

= -1

e, -(-11) + (-15) + (13) - 21

toan lop 6 ma chi

\(1,\\ a,=\left[x^3\left(x-2\right)-4x\left(x-2\right)\right]:\left(x^2-4\right)\\ =x\left(x^2-4\right)\left(x-2\right):\left(x^2-4\right)=x\left(x-2\right)\\ b,=\left(2014-14\right)^2=2000^2=4000000\\ 2,\\ A=2015\cdot2013\cdot\left(2014^2+1\right)\\ A=\left(2014^2-1\right)\left(2014^2+1\right)\\ A=2014^4-1< B=2014^4\)