Giải các phương trình: a, $\frac{x^{2}}{9}$ + $\frac{16}{x^{2}}$ =$\frac{10}{3}$($\frac{x}{3}-$$\frac{4}{x}$) b, $\frac{1}{x(x+2)}$ – $\frac{1}{(x+1)^

0 bình luận về “Giải các phương trình: a, $\frac{x^{2}}{9}$ + $\frac{16}{x^{2}}$ =$\frac{10}{3}$($\frac{x}{3}-$$\frac{4}{x}$) b, $\frac{1}{x(x+2)}$ – $\frac{1}{(x+1)^”

1. Giải thích các bước giải:

   a.Ta có:

   $\dfrac{x^2}{9}+\dfrac{16}{x^2}=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$

   $\to \dfrac{x^2}{9}+2\cdot \dfrac{x}3\cdot \dfrac4x+\dfrac{16}{x^2}-\dfrac83=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$

   $\to (\dfrac{x}3-\dfrac4x)^2-\dfrac83=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$

   $\to (\dfrac{x}3-\dfrac4x)^2-\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)-\dfrac83=0$

   $\to (\dfrac{x}{3}-\dfrac4x+\dfrac23)(\dfrac{x}{3}-\dfrac4x-4)=0$

   $\to \dfrac{x}{3}-\dfrac4x+\dfrac23=0$

   $\to x^2-12+2x=0$

   $\to x=-1\pm\sqrt{13}$

   Hoặc $\dfrac{x}{3}-\dfrac4x-4=0$

   $\to x^2-12-12x=0$

   $\to x=6\pm4\sqrt3$

   b.ĐKXĐ: $x\ne 0,-1,-2$

   Ta có:

   $\dfrac{1}{x(x+2)}-\dfrac{1}{(x+1)^2}=\dfrac{1}{12}$

   $\to \dfrac{1}{x^2+2x}-\dfrac{1}{x^2+2x+1}=\dfrac{1}{12}$

   Đặt $x^2+2x+1=a\to a=(x+1)^2\ge 0$

   $\to \dfrac1{a-1}-\dfrac1a=\dfrac1{12}$

   $\to 12a-12\left(a-1\right)=a\left(a-1\right)$

   $\to 12=a^2-a$

   $\to a^2-a-12=0$

   $\to (a-4)(a+3)=0$

   $\to a-4=0$ vì $a\ge 0\to a+3>0$

   $\to a=4$

   $\to (x+1)^2=4$

   $\to x+1=2\to x=1$ hoặc $x+1=-2\to x=-3$

   c.Ta có:

   $2x^3-3x^2-11x+6=0$

   $\to \left(x+2\right)\left(2x-1\right)\left(x-3\right)=0$

   $\to x\in\{-2,\dfrac12,3\}$

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