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3(x + 1)2 - 3x(x + 2) = 1
<=> 3x2 + 6x + 3 - 3x2 - 6x = 1
<=> 3 = 1 (vô lí)
Vậy phương trình vô nghiệm.
(x - 1)3 - (x + 3)(x2 - 3x + 9) + 3(x2 - 4) = 2
<=> x3 - 3x2 + 3x - 1 - x3 - 27 + 3x2 - 12 = 2
<=> 3x - 40 = 2
<=> 3x = 42
<=> x = 14
Vậy S = { 14 }.
(x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
<=> x3 + 8 - x3 - 2x = 15
<=> - 2x + 8 = 15
<=> - 2x = 7
<=> x = - 7/2
Vậy S = { - 7/2 }.

P/S : Câu 2,3 kết quả bằng bao nhiêu mới tìm được x ?
1.\(\left(2x-7\right)^2-4\left(x-3\right)=5\)
=> \(\left(2x\right)^2-2\cdot2x\cdot7+7^2-4x+12=5\)
=> \(4x^2-28x+49-4x+12=5\)
=> \(4x^2-32x+61=5\)
=> \(4x^2-32x+61-5=0\)
=> \(4x^2-32x+56=0\)
=> \(4\left(x^2-8x+14\right)=0\)
=> \(x^2-8x+14=0\)
=> \(\orbr{\begin{cases}x=4-\sqrt{2}\\x=\sqrt{2}+4\end{cases}}\)
4.\(\left(3x-1\right)^2-6\left(x-1\right)\left(x+1\right)-3x\left(x-2\right)=7\)
=> \(\left(3x\right)^2-2\cdot3x\cdot1+1^2-6\left(x^2-1\right)-3x^2+6x=7\)
=> \(9x^2-6x+1-6x^2+6-3x^2+6x=7\)
=> \(\left(9x^2-6x^2-3x^2\right)+\left(-6x+6x\right)+\left(1+6\right)=7\)
=> 7 = 7(đúng)
5. \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
=> \(x^2+2\cdot x\cdot3+3^2-x\left(x+8\right)+4\left(x+8\right)=1\)
=> x2 + 6x + 9 - x2 - 8x + 4x + 32 = 1
=> (x2 - x2) + (6x - 8x + 4x) + (9 + 32) = 1
=> 2x + 41 = 1
=> 2x = -40
=> x = -20

\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự

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
(x+1-2+x).(x+1+2-x)=4
(2x-1).3=4
6x -3-4=0
6x-7=0
1) (x + 1)2 - (2 - x)2 = 4
<=> x2 + 2x + 1 - x2 + 4x - 4 = 4
<=> (x2 - x2) + (2x + 4x) + (1 - 4) = 4
<=> 6x - 3 = 4
<=> 6x = 4 + 3
<=> 6x = 7
<=> x = 7 : 6 = 7/6
=> x = 7/6
2) (x + 2)2 + x(3 - x) = 3
<=> x2 + 4x + 4 + 3x - x2 = 3
<=> (x2 - x2) + (4x + 3x) + 4 = 3
<=> 7x + 4 = 3
<=> 7x = 3 - 4
<=> 7x = -1
<=> x = -1 : 7 = -1/7
=> x = -1/7