Nhận thấy x = 0 không phải là nghiệm.

Xét x khác 0.Chia hai vế của pt cho x² ta được:

\(x^2-3x-6+\frac{3}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-6=0\)

Đặt \(x-\frac{1}{x}=a\). PT trở thành:

\(a^2-3a-4=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-1\end{matrix}\right.\)

Với a = 4 thì \(x=4+\frac{1}{x}=\frac{4x+1}{x}\Leftrightarrow x^2-4x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{5}\\x=2-\sqrt{5}\end{matrix}\right.\) (nghiệm xấu chút nhưng dễ giải lắm ạ)

Với a = -1 thì \(x=\frac{1}{x}-1=\frac{1-x}{x}\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1+\sqrt{5}}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\) (cái này thì max xấu rồi ;( )

\(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\) (Đkxđ: \(x\ne-7;x\ne\frac{3}{2}\))

\(\Rightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-9x-4x-6x^2-42x-x=7-6\)

\(\Leftrightarrow-56x=1\)

\(\Leftrightarrow x=-\frac{1}{56}\) (t/m đkxđ)

Vậy \(S=\left\{-\frac{1}{56}\right\}\)

ĐKXĐ: x khác -7 và 3/2

Từ đề bài <=> (3x-2)(2x-3) = (6x+1)(x+7)

<=> 6x^2-4x-9x+6 = 6x^2+x+42x+7

<=> -13x+6 = 43x+7

<=> 6-7 = 43x+13x

<=> 56x = -1

<=> x = -1/56 (TM)

Vậy ...

ĐKXĐ:x khác -7;x khác 1,5

=>(3x-2)(2x-3)=(6x+1)(x+7)

=>6x²-4x-9x+6=6x²+x+42x+7

=>6x²-13x+6=6x²+43x+7

=>6x²-6x²-13x-43x+6-7=0

=>-56x-1=0

=>-56x=1

=>x=\(\frac{-1}{56}\)

Ta có

\(\left\{{}\begin{matrix}\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\\\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\) \(\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\) \(\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\) \(\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4}=2\\3-\left(x-1\right)^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

ĐKXĐ: ...

\(\Leftrightarrow\frac{2x}{3x^2-4x+1}-\frac{7x}{3x^2+2x+1}=6\)

\(\Leftrightarrow\frac{2}{3x-4+\frac{1}{x}}-\frac{7}{3x+2+\frac{1}{x}}=6\)

Đặt \(3x-4+\frac{1}{x}=a\)

\(\frac{2}{a}-\frac{7}{a+6}=6\)

\(\Leftrightarrow2\left(a+6\right)-7a=6a\left(a+6\right)\)

\(\Leftrightarrow6a^2+41a-12=0\)

Nghiệm xấu, bạn coi lại đề

GPT

\(\frac{3}{3x^2-4x+1}+\frac{13}{3x^2+2x+1}=\frac{6}{x}\)

a)

Theo bài ra ta có :

\(\left(x+7\right)\left(3x-1\right)-x^2+49=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(x^2-49\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(\left(x-7\right)\left(x+7\right)\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1-x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x+7=0\\2x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-7\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;-7\right\}\)

Chúc bạn học tốt =))

a/

<=>(x+7)(3x-1)-(x^2-7^2)=0

<=>(x+7)(3x-1)-(x-7)(x+7)=0

<=>(x+7)(3x-1-x+7)=0

<=>(x+7)(2x+6)=0

<=>x+7=0 hoặc 2x+6=0

<=>x=-7 2x=-6

<=> x=-3

=>S (-7;-3)

Cho hỏi cái đề câu b là đây hả:

\(\frac{x^2-3x+1}{x^2}-\frac{3}{x}=-4\)

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\)\(\frac{x^2-3x+1}{x^2}-\frac{3x}{x^2}=\frac{-4x^2}{x^2}\)

\(\Leftrightarrow\)\(x^2-3x+1-3x=-4x^2\)

\(\Leftrightarrow x^2+4x^2-6x+1=0\)

\(\Leftrightarrow5x^2-6x+1=0\)

\(\Leftrightarrow5x^2-5x-x+1=0\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\left[\begin{matrix}5x-1=0\Rightarrow x=0,2\\x-1=0\Rightarrow x=1\end{matrix}\right.\)

\(\Leftrightarrow\left(36x^2+84x+48\right)\left(36x^2+84x+49\right)=72\)

\(\Leftrightarrow t\left(t+1\right)=72\) ( với \(t=36x^2+84x+48\) )

\(\Leftrightarrow t^2+t-72=0\Leftrightarrow\left(t-8\right)\left(t+9\right)=0\)

\(\Leftrightarrow t-8=0\) ( do \(t+9=36x^2+84x+49+8=\left(6x+7\right)^2+8>0\forall x\))

\(\Leftrightarrow36x^2+84x+48=8\)

\(\Leftrightarrow\left(6x+7\right)^2=9\Leftrightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{matrix}\right.\) ( TM )

x=\(\dfrac{-2}{3}\)