đặt biểu thức trên =A

ta có: A=2/13.19+2/19.25+2/25.31+...+2/613.619

=(2/13-2/19)+(2/19-2/25)+(2/25-2/31)+...+(2/613-2/619)

=2/13-2/19+2/19-2/25+2/25-2/31+...+2/613-2/619

=2/13+(2/19-2/19)+(2/25-2/25)+(2/31-2/31)+...+(2/613-2/613)-2/619

=2/13-2/619=1212/8047

= 2 .[ \(\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+........+\dfrac{1}{613.619}\)]

= 2 .[\(\dfrac{1}{6}\).(\(\dfrac{1}{13}-\dfrac{1}{19}\))+\(\dfrac{1}{6}\). (\(\dfrac{1}{19}-\dfrac{1}{25}\))+\(\dfrac{1}{6}\). (\(\dfrac{1}{25}-\dfrac{1}{31}\)) + ......+\(\dfrac{1}{6}\). (\(\dfrac{1}{613}-\dfrac{1}{619}\) ) ]

= 2. [\(\dfrac{1}{6}\).( \(\dfrac{1}{13}-\dfrac{1}{19}\)+\(\dfrac{1}{19}-\dfrac{1}{25}\)+\(\dfrac{1}{25}-\dfrac{1}{31}\)+ ......+\(\dfrac{1}{613}-\dfrac{1}{619}\))]

= 2. [\(\dfrac{1}{6}\).\(\dfrac{1}{619}\)]

=2 . \(\dfrac{1}{3714}\)

= \(\dfrac{1}{1857}\)

có j sai mọi người cứ nói

= 2 .[

= 2 .[.()+. ()+. () + ......+. () ]

= 2. [ 1/6+1/6. 1/6.1+ 1/6. 1/6. + ......+ 1/6. 1/6. )]

= 2. [.1/6.]

=2 . [1/6.(1/3 - 1/619]

= 2. [1/6 .616/1857]

= 2.308/5571

=616/5571

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!

\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)

\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)

\(=\dfrac{486}{375}=\dfrac{162}{125}\)

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`

`=(2x+6)/(x-2)`

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)

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

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

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

`=((x-3)^2/(x^2-9)+(4x^2)/(x^2-9)-(x+3)^2/(x^2-9)):((x-2)/(x+3))`

`=(((x-3)^2+4x^2-(x+3)^2)/(x^2-9))*(x+3)/(x-2)`

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

`=(4x(x-3))/((x-3)(x+3))*(x+3)/(x-2)`

`=(4x)/(x+3)*(x+3)/(x-2)`

`=(4x)/(x-2)`

a: Ta có: \(A=\dfrac{1}{2}\)

\(\Leftrightarrow x+2=2x-6\)

\(\Leftrightarrow-x=-8\)

hay x=8

Thay x=8 vào B,ta được:

\(B=-\dfrac{2}{8+2}=-\dfrac{2}{10}=-\dfrac{1}{5}\)

\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)

x trong bài là dấu nhân nha, mik viết nhầm

B=8+3.1/4-1/4+(4:1/2).8

=8+3/4-1/4+8.8

=8+3/4-1/4+64

=35/4-275/4

=-60