a) \(\left(x-3\right)^2+2x-6=0\)

\(\Leftrightarrow x^2-6x+9+2x-6=0\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\) (ĐKXĐ: \(x\ne\pm3\))

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x+3}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2+6x+9-48=x^2-6x+9\)

\(\Leftrightarrow x^2-x^2+6x+6x+9-9-48=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=\dfrac{48}{12}\)

\(\Leftrightarrow x=4\left(tm\right)\)

a: (x-3)^2+2x-6=0

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

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

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

=>x=3 hoặc x=1

b:

ĐKXĐ: x<>3; x<>-3

 \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\)

=>\(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48}{\left(x-3\right)\cdot\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)^2}\)

=>(x+3)^2-48=(x-3)^2

=>x^2+6x+9-48=x^2-6x+9

=>6x-39=-6x+9

=>12x=48

=>x=4(nhận)

\(a,\left(x-3\right)^2-2x+6=0\\ \Leftrightarrow x^2-6x+9-2x+6=0\\ \Leftrightarrow x^2-8x+15=0\\ \Leftrightarrow x^2-3x-5x+15=0\\ \Leftrightarrow x\left(x-3\right)-5\left(x-3\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\\ Vậy:S=\left\{5;3\right\}\\ b,\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{x^2-9}=0\\ \Leftrightarrow\left[\left(x+3-x+3\right)\left(x+3+x-3\right)\right]-48=0\\ \Leftrightarrow6.2x-48=0\\ \Leftrightarrow12x=48\\ \Leftrightarrow x=4\left(TM\right)\\ Vậy:S=\left\{4\right\}\)

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

a: (x+1)^3-x(x-2)^2+x-1=0

=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0

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

=>7x^2=0

=>x=0

b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x=2+1+27+12=39+3=42

=>x=14

a: \(\Leftrightarrow3x\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

Rối quá bn ơi

Bn ko dùng dấu enter để xuống dòng à 

Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao

a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)

(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)

(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)

(=)\(\left(14x+5\right)\left(22x+12\right)=0\)

(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)

1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)

2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)

Vậy........

b)\(\left(x+1\right)^2+2\left(x+1\right)+1=0\)

(=)\(\left(x+1\right)^2+2\left(x+1\right).1+1^2=0\)

(=) \(\left(x+2\right)^2=0\)

(=) \(x+2=0\Rightarrow x=-2\)

Vậy........

c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)

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

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

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

(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)

1)\(x+3=0\Rightarrow x=-3\)

2)\(x-2=0\Rightarrow x=2\)

3) \(x-1=0\Rightarrow x=1\)

Vậy.........

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)