`Parkchaeyoung`

 

Đáp án:

a) $S=\{-20\}$

b) $S=\{-4\}$

c) $S=\left\{1;-\dfrac{7}{6}\right\}$

d) $S=\{1\}$

Giải thích các bước giải:

a)

$\dfrac{2x-5}{x+5}=3\,\,\,(x\ne-5)\\\to 2x-5=3x+15\\⇔3x-2x=-15-5\\⇔x=-20$ (thoả mãn)

Vậy $S=\{-20\}$

b)

$\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\,\,\,(x\ne0)\\⇔\dfrac{x^2-6}{x}-x-\dfrac{3}{2}=0\\⇔\dfrac{2(x^2-6)-2x^2-3x}{2x}=0\\\to 2x^2-12-2x^2-3x=0\\⇔3x=-12\\⇔x=-4$ (thoả mãn)

Vậy $S=\{-4\}$

c)

$\dfrac{5}{3x+2}=2x-1\,\,\,\left(x\ne-\dfrac{2}{3}\right)\\\to (3x+2)(2x-1)=5\\⇔6x^2+x-2=5\\⇔6x^2+x-7=0\\⇔(6x^2-6x)+(7x-7)=0\\⇔6x(x-1)+7(x-1)=0\\⇔(x-1)(6x+7)=0\\⇔\left[\begin{array}{l}x=1\\x=-\dfrac{7}{6}\end{array}\right.$

Vậy $S=\left\{1;-\dfrac{7}{6}\right\}$

d)

$x+\dfrac{1}{x}=\dfrac{1}{x^2}+x^2\,\,\,(x\ne0)$

Đặt $t=x+\dfrac{1}{x}$

$\to t^2=\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}+2\\\to x^2+\dfrac{1}{x^2}=t^2-2$

Phương trình trở thành:

$t=t^2-2\\⇔t^2-t-2=0\\⇔(t^2+t)-(2t+2)=0\\⇔t(t+1)-2(t+1)=0\\⇔(t+1)(t-2)=0\\⇔\left[\begin{array}{l}t=-1\\t=2\end{array}\right.\\⇔\left[\begin{array}{l}x+\dfrac{1}{x}=-1\\x+\dfrac{1}{x}=2\end{array}\right.\\⇔\left[\begin{array}{l}x^2+1=-x\\x^2+1=2x\end{array}\right.\\⇔\left[\begin{array}{l}x^2+x+1=0\\x^2-2x+1=0\end{array}\right.\\⇔\left[\begin{array}{l}\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\(x-1)^2=0\end{array}\right.\\⇔x=1$ (thoả mãn)

Vậy $S=\{1\}$