\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\)

\(ĐKXĐ:x\ne\pm2\)

\(pt\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2-3x+2}{x^2-4}+\frac{3x+6}{x^2-4}\)

\(\Leftrightarrow\frac{9}{x^2-4}=\frac{x^2+8}{x^2-4}\)

\(\Leftrightarrow x^2+8=9\Leftrightarrow x=\pm1\left(tm\right)\)

Vậy pt có 2 nghiệm là 1 và -1

Điều kện :  \(x+2\ne0\) và \(x-2\ne0\Leftrightarrow x=\pm2\)

( Khi đó \(x^2-4=\left(x+2\right)\left(x-2\right)\ne0\) )

\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\)

\(\Leftrightarrow\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)+3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-3x+2+3x+6=9\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)

Vậy tập nghiệm của PT là: \(S=\left\{-1;1\right\}\)

Chúc bạn học tốt !!!

\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

=>\(3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

=>\(3x^2-12x+12=3x^2+3x-9-9x+9\)

=>\(3x^2-12x+12=3x^2-6x\)

=>-6x=-12

=>x=2

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy: S={3;-3}

Bài 1:

Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:

\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)

\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc

\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)

Câu 3: ĐK: \(x\ge0\)

Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)

Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)

\(\Rightarrow2x-4\sqrt{x-1}=0\)

Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)

Câu 3 :

ĐKXĐ : \(x\ge1\)

\(3\left(x^2-x+1\right)=\left(x+\sqrt{x-1}\right)^2\)

\(\Leftrightarrow3\left[x^2-\left(x-1\right)\right]=\left(x+\sqrt{x-1}\right)^2\)

\(\Leftrightarrow3\left(x-\sqrt{x-1}\right)\left(x+\sqrt{x-1}\right)=\left(x+\sqrt{x-1}\right)^2\)

\(\Leftrightarrow\left(x+\sqrt{x-1}\right)\left(x+\sqrt{x-1}-3x+3\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x-1}\right)\left(4\sqrt{x-1}-2x\right)=0\)

Tới đây thì dễ rồi ^^

1) `x^2+4-2(x-1)=(x-2)^2`

`<=>x^2+4-2x+2=x^2-4x+4`

`<=>-2x+2=-4x`

`<=>2x=-2`

`<=>x=-1`

.

2) ĐKXĐ: `x \ne \pm 3`

`(x+3)/(x-3)-(x-1)/(x+3)=(x^2+4x+6)/(x^2-9)`

`<=>(x+3)^2-(x-1)(x-3)=x^2+4x+6`

`<=>x^2+6x+9-x^2+4x-3=x^2+4x+6`

`<=>10x+6=x^2+4x+6`

`<=>x^2-6x=0`

`<=>x(x-6)=0`

`<=>x=0;x=6`

.

3) ĐKXĐ: `x \ne \pm 3`

`(3x-3)/(x^2-9) -1/(x-3 )= (x+1)/(x+3)`

`<=>(3x-3)-(x+3)=(x+1)(x-3)`

`<=> 2x-6=x^2-2x-3`

`<=>x^2-4x+3=0`

`<=>x^2-x-3x+3=0`

`<=>x(x-1)-3(x-1)=0`

`<=>(x-3)(x-1)=0`

`<=> x=3;x=1`

Vậy...

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

 \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

=>\(\left(x+3\right)^2-4=0\)

=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

=>(x-2-3)(x-2+3)=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

PT \(\Leftrightarrow9x^2-6x+1-9x+6=9x^2-18x-27\)

\(\Leftrightarrow9x^2-6x+1-9x+6-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy ...

Ta có: \(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-3x+x-3\right)\)

\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)

\(\Leftrightarrow9x^2-15x+7-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow3x=-34\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy: \(S=\left\{-\dfrac{34}{3}\right\}\)