Đáp án:

$\begin{array}{l}
a)Dkxd:x\# 0\\
x + \dfrac{1}{x} = {x^2} + \dfrac{1}{{{x^2}}}\\
 \Leftrightarrow x + \dfrac{1}{x} = {x^2} + 2.{x^2}.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^2}}} - 2\\
 \Leftrightarrow x + \dfrac{1}{x} = {\left( {x + \dfrac{1}{x}} \right)^2} - 2\\
Đặt:x + \dfrac{1}{x} = t\\
 \Leftrightarrow t = {t^2} - 2\\
 \Leftrightarrow {t^2} - t - 2 = 0\\
 \Leftrightarrow \left( {t - 2} \right)\left( {t + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t - 2 = 0\\
t + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} - 2 = 0\\
x + \dfrac{1}{x} + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x + 1 = 0\\
{x^2} + x + 1 = 0\left( {ktm} \right)
\end{array} \right.\\
 \Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
b)Dkxd:x\# \dfrac{2}{3};x\# \dfrac{3}{2}\\
\dfrac{{3x - 2}}{{x + 7}} = \dfrac{{6x + 1}}{{2x - 3}}\\
 \Leftrightarrow \left( {3x - 2} \right).\left( {2x - 3} \right) = \left( {6x + 1} \right)\left( {x + 7} \right)\\
 \Leftrightarrow 6{x^2} - 9x - 4x + 6 = 6{x^2} + 42x + 7x + 7\\
 \Leftrightarrow 62x =  - 1\\
 \Leftrightarrow x =  - \dfrac{1}{{62}}\left( {tmdk} \right)\\
Vậy\,x =  - \dfrac{1}{{62}}\\
c)Dkxd:x\#  - 3\\
\dfrac{{2x - 2{x^2}}}{{x + 3}} = \dfrac{{4x}}{{x + 3}} + \dfrac{2}{7}\\
 \Leftrightarrow \dfrac{{7\left( {2x - 2{x^2}} \right)}}{{7\left( {x + 3} \right)}} = \dfrac{{7\left( {4x + 2} \right) + 2.\left( {x + 3} \right)}}{{7\left( {x + 3} \right)}}\\
 \Leftrightarrow 14x - 14{x^2} = 28x + 14 + 2x + 6\\
 \Leftrightarrow 14{x^2} + 16x + 20 = 0\\
 \Leftrightarrow 7{x^2} + 8x + 10 = 0\left( {ktm} \right)
\end{array}$

Vậy pt vô nghiệm

Đáp án:

Giải thích các bước giải:

b║

⇔(3x−2)(2x−3)=(6x+1)(x+7)

⇔6x²−9x−4x+6=6x²+42x+x+7

⇔6x²−13x+6=6x²+42x+x+7

⇔6x²−13x+6=6x²+43x+7

⇔−13x+6=43x+7

⇔6−7=56x

⇒x=−$\frac{1}{56}$ 

a║

⇔1+$\frac{1}{x}$ = $\frac{x²+1}{x²}$

⇔1+$\frac{1}{x}$=1+$\frac{1}{x²}$

⇔​$\frac{1}{x}$ =$\frac{1}{x}$ 

⇔$x^{2}$ =x

⇔$x^{2}$ -x=0

⇒x=1

c║

ĐKXĐ x∦ -3

2-$\frac{2x²}{x+3}$ =$\frac{4x}{x+3}$ +$\frac{2}{7}$

$\frac{14x (x+3) -14x²}{7(x+3)}$ =$\frac{28+2(x+3)}{7(x+3)}$

14x² +42x -14x²=28x+2x+6x

42x-28x-2x-6

12x=6

x=$\frac{1}{2}$

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