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a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

a) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x( x - 3 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 - 18x
= x3 - 6x - 8 - x3 + x
= -5x - 8
b) ( x + 1 )3 - ( x - 1 )3 - 6( x - 1 )2
= x3 + 3x2 + 3x + 1 - ( x3 - 3x2 + 3x - 1 ) - 6( x2 - 2x + 1 )
= x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6
= 12x - 4
c) ( 2x + 1 )( 4x2 - 2x + 1 ) + ( 2 - 3x )( 4 + 6x + 9x2 ) - 9
= ( 2x )3 + 13 + 23 - ( 3x )3 - 9
= 8x3 + 1 + 8 - 27x3 - 9
= -19x3
d) ( x + 1 )3 + ( x - 1 )3 + x3 - 3x( x - 1 )( x + 1 )
= x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 + x3 - 3x( x2 - 1 )
= 3x3 + 6x - 3x2 + 3x
= 9x

1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64

a: \(\Leftrightarrow\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=4\)
\(\Leftrightarrow\left(3x+1-3x-5\right)^2=4\)
=>16=4(vô lý)
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2