\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)

= \(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)

= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

= \(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)

= \(2.\frac{49}{100}\)

= \(\frac{49}{50}\)

là sao ????=))

giữa các phân số là cộng hay trừ vậy???

\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\) 

\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\) 

\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\dfrac{7}{18}:\dfrac{1}{2}\) 

\(=\dfrac{7}{9}\)

2 x - 1 + 2 x + 3 x 2 + x + 1 = 2 x - 1 2 x + 1 x 3 - 1     Đ K X Đ : x ≠ 1 ⇔ 2 x 2 + x + 1 x 3 - 1 + 2 x + 3 x - 1 x 3 - 1 = 2 x - 1 2 x + 1 x 3 - 1

⇔ 2( x 2  + x + 1) + (2x + 3)(x – 1) = (2x – 1)(2x + 1)

⇔ 2 x 2 + 2x + 2 + 2 x 2  – 2x + 3x – 3 = 4 x 2  – 1

⇔ 2 x 2  + 2 x 2  – 4 x 2  + 2x – 2x + 3x = -1 – 2 + 3

⇔ 3x = 0 ⇔ x = 0 (thỏa mãn)

Vậy phương trình có nghiệm x = 0.

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

⇔ 4x - 10 = 2 - x

⇔ 4x + x = 2 + 10 ⇔ 5x = 12 ⇔ x = 12/5

Vậy: S = {12/5}

b) (3x + 1) = (3x + 1)2

⇔ (3x + 1)2 - (3x + 1) = 0

⇔ (3x + 1)[(3x + 1) - 1] = 0

ĐKXĐ:x ≠ 1

Quy đồng mẫu hai vế của phương trình ta được:

Khử mẫu hai vế, ta được:

(2x + 3)(x - 1) + 2(x2 + x + 1) = 4x2 - 1

⇔ 2x2 + x - 3 + 2x2 + 2x + 2 = 4x2 - 1

⇔ 3x - 1 = -1

⇔ 3x = 0 ⇔ x = 0 (thỏa mãn điều kiện)

Vậy: S = {0}

a,A =  \(\dfrac{3x^2+6xy}{6x^2}\)  (đk  \(x\) ≠ 0)

  A = \(\dfrac{3x.\left(x+2y\right)}{6x^2}\)

 A = \(\dfrac{x+2y}{2x}\) 

b,B =  \(\dfrac{2x^2-x^3}{x^2-4}\) (đk \(x\)²  - 4 ≠ 0 ⇒ \(x\) ≠ \(\pm\) 2)

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

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

   B =  \(\dfrac{-x^2}{x+2}\)

c, \(\dfrac{x+1}{x^3+1}\)  (đk \(x^3\) + 1 ≠ 0 ⇒ \(x\) ≠ -1)

   C = \(\dfrac{x+1}{\left(x+1\right).\left(x^2-x+1\right)}\)

  C = \(\dfrac{1}{\left(x^2-x+1\right)}\)

a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

= ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) +...+ ( 1/2001 - 1/2003) + (1/2003 - 1/2005)

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005