Ta có:

\(\frac{1}{2x2}<\frac{1}{1.2}\)

\(\frac{1}{3x3}<\frac{1}{2.3}\)

\(...\)

\(\frac{1}{2015x2015}<\frac{1}{2014x2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{2014x2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2015}<1\)

\(\Rightarrow\)Đpcm

\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)

\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(A>1-\frac{1}{2015}\)

Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)

a: \(A\left(x\right)=5x^2-2x^3+4x^5+3x^3-3x^2+2x-1\)

\(=4x^5+\left(3x^3-2x^3\right)+\left(5x^2-3x^2\right)+2x-1\)

\(=4x^5+x^3+2x^2+2x-1\)

Bậc của A(x) là 5

\(B\left(x\right)=-x^5+2x^3-3x^5-2x^2-3x^3+3x-5\)

\(=\left(-x^5-3x^5\right)+\left(2x^3-3x^3\right)-2x^2+3x-5\)

\(=-4x^5-x^3-2x^2+3x-5\)

Bậc của B(x) là 5

b: C(x)=A(x)+B(x)

\(=4x^5+x^3+2x^2+2x-1-4x^5-x^3-2x^2+3x-5\)

=5x-6

c: C(-1)=5*(-1)-6

=-5-6

=-11

d: Đặt C(x)=0

=>5x-6=0

=>5x=6

=>x=1,2

A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+98.99(100-97)   "." la dau nhan

A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+98.99.100-97.98.99

A=1.2.3+98.99.100

A= 970206

Ta có : B = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3B = 0.1.2 + 1.2.3 - 1.2.3 + ...... + 98.99.100

=> 3B = 98.99.100

=> B = \(\frac{98.99.100}{3}\)  = 323400

a) M = 4 x 5 y   –   6 x 3 y 2   +   10 x 3 y 2 z .

b) N =  x 4 y 3   –   2 x 2 y 4   +   x 2 y 3

trả lời

các gt ll là

4

9

16

30

2

20

hok tốt

2x2=4

3x3=9

4x4=16

5x6=30

1x2= 2

4x5= 20

                                                                     \(Giải\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)

      \(A=0+0+0+...+0+0\)

      \(\Rightarrow A=0\)   

\(a.\)\(A< 1\)

b.   \(A< \frac{3}{4}\)