( m : 1 - m . 1 ) : ( m . 2005 + m + 1 )

= ( m - m ) : ( m . 2005 + m + 1 )

= 0 : ( m . 2005 + m + 1 )

= 0

       \(\left(m:1-m\times1\right):\left(m\times2005+m+1\right)\)

\(=\left(m\times\frac{1}{1}-m\times1\right):\left(m\times2006+1\right)\)

\(=\left(m\times1-m\times1\right):\left(m\times2006+1\right)\)

\(=0:\left(m\times2006+1\right)\)

\(=0\)

(m : 1 - m x 1) : (m x 2005 + m + 1)

= (m - m x 1) : (m x 2005 + m + 1)

= (m - m) : (m x 2005 + m + 1)

= 0 : (m x 2005 + m + 1)

= 0

\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)

\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)

Rút gọn các thừa số ở tử và mẫu ta được:

\(B=\frac{1}{2004}\)

                                                                              Đ/S:\(\frac{1}{2004}\)

Ta có:

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)

Đơn giản hết sẽ là:

\(=\frac{1}{2004}\)

B = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

B = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x.........x\frac{2002}{2003}x\frac{2003}{2004}\)

B = \(\frac{1x2x3x4x..........x2002x2003}{2x3x4x5x.....x2003x2004}\)

Sau khi trực tiêu các số thì ta có:

B = \(\frac{1}{2004}\)

-.-? lớp 6 

Có \(\left(5x-7\right)\left(2x+3\right)-\left(7x+2\right)\left(x-4\right)\)

\(=10x^2+15x-14x-21-7x^2+28x-2x+8\)

\(=\left(10x^2-7x^2\right)+\left(15x-14x+28x-2x\right)+\left(-21+8\right)\)

\(=3x^2+27x-13\)

Thay x = 1/2 vào biểu thức đã rút gọn 

\(\Rightarrow3\left(\frac{1}{2}\right)^2+27.\frac{1}{2}-13=\frac{5}{4}\)

Vậy giá trị biểu thức là 5/4

(5x-7)(2x+3)-(7x+2)(x-4)

=10x²+15x-14x-21-(7x²-7x+2x-8)

=10x²-x-21-7x²+7x-2x+8

=3x²+4x-13

Thay x=1/2 vào BT ta được:

3*(1/2)²+4*1/2-13

=3/4+2-13

=3/4-11

=-41/4

diễn giải ra nhé

X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000

\(B=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(\Rightarrow B=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(\Rightarrow B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)

\(\Rightarrow B=\frac{3.4.5...100}{2.3.4...99}\)

\(\Rightarrow B=\frac{100}{2}\)

\(\Rightarrow B=50\)

Vậy \(B=50\)

a, 2006 x 2004 - \(\frac{2}{1995}\) + 2004 x 2005 = 8038043,999

b, 2006 x 125 + \(\frac{1000}{126}\) x 2006 - 1006 = 265664,6349

c, A = 1991 x 1999

=> A = ( 1995 - 4 ) x ( 1995 + 4 )

     A = 1995 x ( 1995 + 4 ) - 4 x ( 1995 + 4 )

     A = 1995 x 1995 + 1995 x 4 - ( 4 x 1995 + 4 x 4 )

     A = 1995 x 1995 - 4 x 4

mà B = 1995 x 1995

Vậy A < B

d, Gọi giá trị biểu thức là C

C =  \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\) 

C x 2 = \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)

C x 2 = \(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)

Vậy C x 2 - C = \(\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)\)

                  C = \(\frac{2}{3}-\frac{1}{96}\) ( vì phân số nào có ở số bị trừ cũng có ở số trừ thì trừ hết rồi nên không còn )

                  C = \(\frac{21}{32}\)

A=1991x1999=(1995-4)1999=1995x1999-4x1999

B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A

vậy A<B

\(\left(1-\frac{1}{9}\right)x\left(1-\frac{1}{16}\right)x\left(1-\frac{1}{21}\right)x.........x\left(1-\frac{1}{210}\right)\)

= \(\frac{8}{9}x\frac{15}{16}x\frac{20}{21}x.........x\frac{209}{210}\)

Sau đó bạn tìm ra quy luật thôi

\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\frac{-3}{8}+\frac{1}{2}\)(tự quy đồng)

\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\frac{1}{8}\)

\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow x+1=3\)

\(\Leftrightarrow x=3-1\)

\(\Leftrightarrow x=2\)

ta có ( 1/2 )^{x + 1} = 1/2 suy ra x +1 = 1 Nên x = 0