Phần tự luận (7 điểm) SVIP

Hướng dẫn giải:

a) $\dfrac{2y-1}{y}-\dfrac{2x+1}{x}$

$=\dfrac{x(2y-1)}{xy} - \dfrac{y(2x+1)}{xy}$

$=\dfrac{2xy - x - 2xy - y}{xy} = \dfrac{-x-y}{xy}$.

b) $\dfrac{2x}{3} \, : \, \dfrac{5}{6x^2}$

$\dfrac{2x}{3} \, : \, \dfrac{5}{6x^2}=\dfrac{2x}{3}.\dfrac{6x^2}{5}$

$=\dfrac{4x^3}{5}$.

Hướng dẫn giải:

a)  Với $x \ne \dfrac{1}{3}$, $x \ne -\dfrac{1}{3}$. ta có:

$P= \left(\dfrac{2x}{3x+1}-1 \right) \, : \, \left(1-\dfrac{8x^2}{9x^2-1} \right)$

$= \dfrac{2x - 3x - 1}{3x+1} \, : \, \dfrac{9x^2 - 1 - 8x^2}{9x^2-1}$

$=\dfrac{-(x+1)}{3x+1}.\dfrac{9x^2-1}{x^2 - 1}$

$=\dfrac{-(x+1)}{3x+1}.\dfrac{(3x+1)(3x-1)}{(x+1)(x-1)}$

$=\dfrac{1-3x}{x-1}$.

b) Thay $x=2$ vào biểu thức ta có:

     $P=\dfrac{1-3.2}{2-1}=-5$.

Hướng dẫn giải:

$\dfrac{x+1}{3}=\dfrac{2x+5}{5}$

$\dfrac{5(x+1)}{15}=\dfrac{3(2x+5)}{15}$

$5x + 5 = 6x + 15$

$5x - 6x = 15 - 5$

$-x = 10$

$x = -10$.

Vậy phương trình có tập nghiệm $S = \left\{ -10 \right\}$.

Hướng dẫn giải:

$AB$ // $DE$.

Theo hệ quả của định lí Thalès ta có:

$\dfrac{CA}{CE}=\dfrac{CB}{CD}=\dfrac{AB}{DE}=\dfrac{5}{15}=\dfrac{1}{3}$

Hay:

⚡$\dfrac{CB}{CD}=\dfrac{1}{3}$ suy ra $\dfrac{x}{7,2}=\dfrac{1}{3}$.

Vậy $x=\dfrac{7,2 . 1}{3}=2,4 $

⚡$\dfrac{CA}{CE}=\dfrac{1}{3}$ suy ra $\dfrac{3}{y}=\dfrac{1}{3}$

Vậy $y=\dfrac{3.3}{1}=9$.

Hướng dẫn giải:

$\dfrac{2x- 50}{50} + \dfrac{2x - 51}{49} + \dfrac{2x - 52}{48} + \dfrac{2x - 53}{47} + \dfrac{2x-200}{25}=0 $

$\dfrac{2x- 50}{50} + \dfrac{2x - 51}{49} + \dfrac{2x - 52}{48} + \dfrac{2x - 53}{47}+\dfrac{2x-100}{25}+\dfrac{-100}{25}=0$

$\dfrac{2x- 50}{50}+\dfrac{2x - 51}{49}+\dfrac{2x - 52}{48}+\dfrac{2x - 53}{47}+\dfrac{2x-100}{25}+(-4)=0$

$\dfrac{2x- 50}{50}-1+\dfrac{2x - 51}{49}-1+\dfrac{2x - 52}{48}-1+\dfrac{2x - 53}{47}-1+\dfrac{2x-100}{25}=0$

$\dfrac{2x-100}{50}+\dfrac{2x - 100}{49}+\dfrac{2x - 100}{48}+\dfrac{2x - 100}{47}+\dfrac{2x-100}{25}=0$

$(2x-100).(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25})=0$

$2x-100=0$ (Do $\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\ne 0$)

$x=50$.

Vậy phương trình đã cho có nghiệm $x = 50$.