pt đã cho \(\Leftrightarrow \frac{2 x - 50}{50} - 1 + \frac{2 x - 51}{49} - 1 + \frac{2 x - 52}{48} - 1 + \frac{2 x - 53}{47} - 1 + \frac{2 x - 200}{25} + 4 = 0\)

\(\Leftrightarrow \frac{2 x - 50 - 50}{50} + \frac{2 x - 51 - 49}{49} + \frac{2 x - 52 - 48}{48} + \frac{2 x - 53 - 47}{47} + \frac{2 x - 200 + 100}{25} = 0\)

\(\Leftrightarrow \frac{2 x - 100}{50} + \frac{2 x - 100}{49} + \frac{2 x - 100}{48} + \frac{2 x - 100}{47} + \frac{2 x - 100}{25} = 0\)

\(\Leftrightarrow \left(\right. 2 x - 100 \left.\right) \left(\right. \frac{1}{50} + \frac{1}{49} + \frac{1}{48} + \frac{1}{47} + \frac{1}{25} \left.\right) = 0\)

\(\Leftrightarrow 2 x - 100 = 0\) (vì \(\frac{1}{50} + \frac{1}{49} + \frac{1}{48} + \frac{1}{47} + \frac{1}{25} > 0\))

\(\Leftrightarrow x = 50\)

Vậy pt đã cho có tập nghiệm 

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = BC/CD

x = BC = AB.CD : DE

x = BC = 5.7,2 : 15 = 2,4

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = AC/CE

y = CE = AC.DE : AB

= 3.15 : 7,2

= 6,25

=> (x+1) . 5 = (2x + 5) . 3 

     5x + 5  = 6x + 15

     5x + 6x = 15-5

           11x =10

              x = 11\10

câu a

\(\left(\right. \frac{2 x}{3 x + 1} - 1 \left.\right) : \left(\right. 1 - \frac{8 x^{2}}{9 x^{2} - 1} \left.\right) = \left(\right. \frac{2 x}{3 x + 1} - \frac{3 x + 1}{3 x + 1} \left.\right) : \left(\right. \frac{9 x^{2} - 1}{9 x^{2} - 1} - \frac{8 x^{2}}{9 x^{2} - 1} \left.\right) = \left(\right. \frac{2 x}{3 x + 1} - \frac{3 x + 1}{3 x + 1} \left.\right) : \left(\right. \frac{9 x^{2} - 1}{\left(\right. 3 x - 1 \left.\right) \left(\right. 3 x + 1 \left.\right)} - \frac{8 x^{2}}{\left(\right. 3 x - 1 \left.\right) \left(\right. 3 x + 1 \left.\right)} \left.\right) = \left(\right. \frac{2 x - 3 x - 1}{3 x + 1} \left.\right) : \left(\right. \frac{9 x^{2} - 1 - 8 x^{2}}{\left(\right. 3 x - 1 \left.\right) \left(\right. 3 x + 1 \left.\right)} \left.\right)\)

\(= \left(\right. \frac{- x - 1}{3 x + 1} \left.\right) : \left(\right. \frac{x^{2} - 1}{\left(\right. 3 x - 1 \left.\right) \left(\right. 3 x + 1 \left.\right)} \left.\right) = \frac{- x - 1}{3 x + 1} \cdot \frac{\left(\right. 3 x - 1 \left.\right) \left(\right. 3 x + 1 \left.\right)}{x^{2} - 1}\)

\(= \frac{- \left(\right. x + 1 \left.\right) \cdot \left(\right. 3 x - 1 \left.\right) \cdot \left(\right. 3 x + 1 \left.\right)}{\left(\right. 3 x + 1 \left.\right) \cdot \left(\right. x - 1 \left.\right) \cdot \left(\right. x + 1 \left.\right)} = \frac{- 3 x + 1}{x - 1}\)

câu b

thay \(x = 2\) vào P ta được

\(\frac{- 3 \cdot 2 + 1}{2 - 1} = \frac{- 6 + 1}{1} = - 5\)

vậy \(P = 5\) khi \(x = 2\)

câu a) 

\(\frac{2 y - 1}{y} - \frac{2 x + 1}{x} = \frac{2 x y - x}{x y} - \frac{2 x y + y}{x y} = \frac{2 x y - x - 2 x y - y}{x y} = \frac{- x - y}{x y}\)

câu b) 

\(\frac{2 x}{3} : \frac{5}{6 x^{2}} = \frac{2 x}{3} \cdot \frac{6 x^{2}}{5} = \frac{2 x \cdot 6 x^{2}}{3 \cdot 5} = \frac{12 x^{3}}{15} = \frac{4 x^{3}}{5}\)

\(\frac{2 x}{3} : \frac{5}{6 x^{2}} = \frac{2 x}{3} \cdot \frac{6 x^{2}}{5} = \frac{2 x \cdot 6 x^{2}}{3 \cdot 5} = \frac{12 x^{3}}{15} = \frac{4 x^{3}}{5}\)

câu a

\(\frac{3 x + 15}{x^{2} - 9} + \frac{1}{x + 3} - \frac{2}{x - 3} = \frac{3 \cdot \left(\right. x + 5 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} + \frac{1}{x + 3} - \frac{2}{x - 3} = \frac{3 \cdot \left(\right. x + 5 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} + \frac{x - 3}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} - \frac{2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)}\)\(= \frac{3 \cdot \left(\right. x + 5 \left.\right) + x - 3 - 2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} = \frac{3 x + 15 + x - 3 - 2 x - 6}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} = \frac{2 x + 6}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} = \frac{2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} = \frac{2}{x - 3}\)

câu b

để \(\frac{2}{x - 3} = \frac{2}{3}\) thì \(x - 3 = 3\)

\(\Rightarrow x = 3 + 3 = 6\)

vậy  \(x = 6\) thì \(A = \frac{2}{3}\)