\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)

\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)

\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)

\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)

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

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

=>x=1 hoặc x=-5

b: =>(x-1)^2=25

=>x-1=5 hoặc x-1=-5

=>x=-4 hoặc x=6

c: =>25x^2+10x+1-25x^2+9=30

=>10x+10=30

=>x+1=3

=>x=2

d: =>x^3-1-x(x^2-4)=5

=>x^3-1-x^3+4x=5

=>4x=6

=>x=3/2

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

a/ \(\left(x+2\right)^2-9=0\)

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

<=> \(\left(x-1\right)\left(x+5\right)=0\)

<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

b/ \(x^2-2x+1=25\)

<=> \(\left(x-1\right)^2=25\)

<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

a) (x+2)²=0

 ==> x+2=0

 ==> x=0-2

==> x=-2

a) (x+2)² - 9 = 0

(x+2)² = 9 = 3² = (-3)²

=> x + 2 = 3

x = 1

=> x+2 = -3

x = - 5

KL: x = 1 hoặc x = -5

b) x² - 2x + 1 = 25

x² - x -x + 1 =25

x.(x-1) - (x -1) = 25

(x-1).(x-1)=25

(x-1)² = 25 = 5² = (-5)²

=> x - 1 = 5

x= 6

=> x - 1 = -5

x = -4

KL: x = 6 hoặc x= -4

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

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

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

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

b) x^2 - 4x + 4 -( x +3) ( x- 3) = 0

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x^2-9\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2-9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x^2=9\end{matrix}\right.\)

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

roois vãi

-Đăng tách câu hỏi bạn nhé.

rối thế bn

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

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

\(=-6x-3\)

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

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

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

<=>  \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)

Vậy...