a) A = 2 + 4 + 6 + 8 + ... + 1000

Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )

               = (1000 + 2) . 500 : 2 = 250500

c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

A = (1000.1) - 999 = 1

B = 100

C: Nhận xét: 16 - 2x8 = 16 - 16 = 0 mà số nào nhân với 0 cũng bằng 0 => C = 0

A:1

B:100

C:0

a: \(\left(1-\frac12\right)\times\left(1-\frac13\right)\times\ldots\times\left(1-\frac{1}{99}\right)\)

\(=\frac12\times\frac23\times\ldots\times\frac{98}{99}\)

\(=\frac{1}{99}\)

b: \(\left(x+\frac12\right)+\left(x+\frac16\right)+\left(x+\frac{1}{12}\right)+\left(x+\frac{1}{20}\right)+\cdots+\left(x+\frac{1}{90}\right)=\frac{99}{10}\)

=>\(\left(x+\frac{1}{1\times2}\right)+\left(x+\frac{1}{2\times3}\right)+\cdots+\left(x+\frac{1}{9\times10}\right)=\frac{99}{10}\)

=>\(9\times x+1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x+1-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x=\frac{99}{10}+\frac{1}{10}-1=\frac{100}{10}-1=10-1=9\)

=>x=1

\(\left(x-99\right)^{1000}+\left(x-100\right)^{2000}=1\)

+) Với \(x=99\)\(;\)\(x=100\) thì \(VT=1\) hay \(x=99\)\(;\)\(x=100\) là nghiệm của pt 

+) Với \(x< 99\) thì \(\left(x-99\right)^{1000}>0\)\(;\)\(\left(x-100\right)^{2000}>1\)

\(\Rightarrow\)\(\left(x-99\right)^{1000}+\left(x-100\right)^{2000}>1\) ( pt vô nghiệm ) 

+) Với \(x>100\) thì \(\left(x-99\right)^{1000}>1\)\(;\)\(\left(x-100\right)^{2000}>0\)

\(\Rightarrow\)\(\left(x-99\right)^{1000}+\left(x-100\right)^{2000}>1\) ( pt vô nghiệm ) 

+) Với \(99< x< 100\) thì \(0< x-99< 1\)\(;\)\(-1< x-100< 0\)

\(\Rightarrow\)\(\left(x-99\right)^{1000}< \left|x-99\right|=x-99\)\(;\)\(\left(x-100\right)^{2000}< \left|x-100\right|=100-x\)

\(\Rightarrow\)\(\left(x-99\right)^{1000}+\left(x-100\right)^{2000}< x-99+100-x=1\) ( pt vô nghiệm ) 

Vậy nghiệm của phương trình là \(x=99\) hoặc \(x=100\)

Chúc bạn học tốt ~ 

(125 x 4 - 1000 : 2) x ( 1 x2 x 3 x... x 99 x 100)

=(500 - 500 ) x ( 1 x2 x 3 x... x 99 x 100)

=0 x ( 1 x2 x 3 x... x 99 x 100)

=0

( 125 x 4 - 1000 : 2 ) x ( 1 x 2 x 3 x ... x 99 x 100 )

Đặt A = 125 x 4 - 1000 : 2

      B = 1 x 2 x 3 x ... x 99 x 100

A = 125 x 4 - 1000 : 2

A = 500 - 500

A = 0

Thay vào , ta có :

0 x ( 1 x 2 x 3 x ... x 99 x 100 )

= 0 

1) \(A=\frac{2x+1}{x^2+2}\)

\(=\frac{\frac{1}{2}\left(x^2+4x+4\right)-\frac{1}{2}\left(x^2+2\right)}{x^2+2}\)

\(=\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}-\frac{1}{2}\ge-\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy GTNN của \(A=-\frac{1}{2}\)khi x = -2 

(x+1)+(x+3)+...+(x+99)=1000

=>50x+1+3+...+99=1000

=>50x+\(\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}\)=1000

=>50x+2500=1000

=>50x=1000-2500=-1500

=>x=-1500:50=-30.

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=1000\)

\(\left(x+x+x+...+x\right)+\left(1+3+5+7+9+...+99\right)\)

\(\left(x.50\right)+2500\)=1000

\(x.50=1000-2500\)

\(x.50=-1500\)

\(x=-1500:50=-30\)