b) A=(X-1,75):0,01

<=>41,25=(X-1,75):0,01

<=> X-1,75=0,4125

<=> X=2,1625

A=(1-0,75).80.X

<=>41,25=0,25.80.X

<=>X=2,0625

Học tốt

a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

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

b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{10}{3}\)

\(=\dfrac{3}{5}\)

ai trả lời đúng thì mình sẽ tick cho

\(A=x^5-5x^4+5x^3-5x^2+5x-6\)

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

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

\(=-2\)

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2003}{2004}\)

\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2003}{2\cdot3\cdot4\cdot...\cdot2004}\)

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

B=(1-1/2)x(1-1/3)x(1-1/4)x(1-1/5)...(1-1/2003)x(1-1/2004)

B=1/2x2/3x3/4x4/5x...x2002/2003x2003/2004

B=1/2004

A)\(a=3\frac{3}{4}\)

suy ra \(\frac{\left(3\frac{3}{4}-1,75\right):0,01}{\left(1-0,75\right)x80}\)

        = 10

b)     nếu A = 41,25 thì ta có \(\frac{\left(a-1,75\right):0,01}{\left(1-0,75\right)x80}=41,25\)

                               suy ra ( a -1,75) : 0,01: ( 1-0,75) x 80 = 41.25

                                         ( a - 1, 75 ) : 0,01 : 0,25 = 41,25 : 80

                                        ( a - 1, 75 ) : 0,01 : 0,25  = 33/64

                                         a - 1,75                         = 33 / 64 x 0,25 x 0,01

                                         a - 1,75                         = 33/25600

                                        a                                    = 33/ 25600+ 1,75

                                       a                                     = 1,751289063 

          = 

b) Ta có: 

\(A=\frac{\left(a-1,75\right)\div0,01}{\left(1-0,75\right)\times80}=41,25\)

\(\frac{\left(a-1,75\right)\div0,01}{0,25\times80}=41,25\)

\(\left(a-1,75\right)\div0,01\div20=41,25\)

\(\left(a-1,75\right)\div0,01=41,25\times20\)

\(\left(a-1,75\right)\div0,01=825\)

\(\left(a-1,75\right)=825\times0,01\)

\(a-1,75=8,25\)

 \(a=8,25+1,75\)

\(a=10\)

Vậy a phải bằng 10 thì biểu thức A có giá trị là 41,25 

\(B=\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{2003}{2004}\)

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

\(B=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times....\left(1-\dfrac{1}{2003}\right)\times\left(1-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\times\dfrac{2}{3}\times....\times\dfrac{2002}{2003}\times\dfrac{2003}{2004}\) 

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

\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\cdot\left(1-\dfrac{1}{2003}\right)\cdot\left(1-\dfrac{1}{2004}\right)\)

\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\cdot\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{2003}{2003}-\dfrac{1}{2003}\right)\cdot\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)

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

\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot2003}{2\cdot3\cdot4\cdot5\cdot...\cdot2004}\)

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

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)

Có vài bước mình làm tắc á nha :>