Giải câu 2 trang 57 sách toán VNEN lớp 9 tập 2

a) $\frac{4x}{x + 2} = \frac{x+1}{x-2}$ (ĐK: $x \neq \pm 2$)

$\Leftrightarrow 4x(x-2)-(x+1)(x+2) = 0$

$\Leftrightarrow 4x^2 -8x-x^2-3x-2=0$

$\Leftrightarrow 3x^2-11x-2=0$

$\Delta = (-11)^2-4\times 3\times (-2) = 145$

$\Rightarrow x_1 = \frac{11+\sqrt{145}}{6} (tmdk); \;\;x_2 = \frac{11-\sqrt{145}}{6}$ (tmđk)

b) $\frac{2x-1}{x} +3 =\frac{x+3}{2x-1}$ (ĐK: $x \neq 0; \; x \neq \frac{1}{2}$)

$\Leftrightarrow (2x-1)^2 + 3x(2x-1)-(x+3)x = 0$

$\Leftrightarrow 9x^2-10x+1 = 0$ (2)

Phương trình (2) có $a+b+c = 0$

$\Rightarrow x_1 = 1 (tmdk);\;x_2 = \frac{1}{9} (tmdk)$

c) $\frac{x-2}{x}+\frac{x}{x-1}-\frac{11}{6} = 0$ (ĐK: $x \neq 0;\; x\neq 1$)

$\Leftrightarrow 6(x-2)(x-1) + 6x^2 -11x(x-1) = 0$

$\Leftrightarrow x^2-7x+12=0$ (3)

$\Delta = (-7)^2 - 4\times 1\times 12 = 1$

$\Rightarrow x_1 = 4 (tmdk);\;x_2 = 3 (tmdk)$

d) $\frac{2x}{x-2}-\frac{5}{x-3}=\frac{5}{x^2-5x+6}$ (ĐK: $x\neq 2;\; 3$

$\Leftrightarrow 2x(x-3) - 5(x-2) - 5=0$

$\Leftrightarrow 2x^2-11x+5 = 0$

$\Delta = (-11)^2 - 4\times 2\times 5 = 81 \Rightarrow \sqrt{\Delta} =9$

$\Rightarrow x_1 = 5 (tmdk); \; x_2 = \frac{1}{2} (tmdk)$

e) $\frac{1}{3x^2-27} +\frac{3}{4}=1+\frac{1}{x-3}$ (ĐK: $x\neq \pm 3$)

$\Leftrightarrow 4 + 3(3x^2 - 27) - 4(3x^2 - 27) - 12(x+3) = 0$

$\Leftrightarrow 4 + 9x^2 -81 -12x^2+108-12x-36=0$

$\Leftrightarrow 3x^2+12x+5 = 0$

$\Delta' = 6^2 - 3\times 5 = 21$

$\Rightarrow x_1 = \frac{-6+\sqrt{21}}{3} (tmdk);\;x_2 = \frac{-6-\sqrt{21}}{3}$