Bài 1: Tìm x
a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=0\)
\(\Leftrightarrow-12x-24=0\)
\(\Leftrightarrow-12x=24\)
hay x=-2
Vậy: x=-2
b) Ta có: \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)-11\left(x-1\right)\left(x+1\right)-6=0\)
\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11\left(x^2-1\right)-6=0\)
\(\Leftrightarrow11x^2+6x+12-11x^2+11=0\)
\(\Leftrightarrow6x+23=0\)
\(\Leftrightarrow6x=-23\)
hay \(x=-\frac{23}{6}\)
Vậy: \(x=-\frac{23}{6}\)
c) Ta có: \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
d) Ta có: \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Vậy: x=-3
a) (2x + 1)2 - 4(x + 2)2 = 9
4x2 + 4x + 1 - 4(x2 + 4x + 4) = 9
4x2 + 4x + 1 - 4x2 - 16x - 16 = 9
-12x - 15 = 9
-12x = 9 + 15
-12x = 24
x = 12 : (-2)
x = -2
b) (3x - 1)2 + 2(x + 3)2 + 11(x + 1)(1 - x) = 6
9x2 - 6x + 1 + 2(x2 + 6x + 9) - 11(x + 1)(x - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11(x2 - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11x2 + 11 = 6
6x + 30 = 6
6x = 6 - 30
6x = -24
x = -24 : 6
x = -4
c) 8x3 - 12x2 + 6x - 1 = 0
(2x)3 - 3.(2x)2.1 + 3.2x.12 - 13 = 0
(2x - 1)3 = 0
2x - 1 = 0
2x = 1
x = 1/2
d) x3 + 9x2 + 27x + 27 = 0
x3 + 3.x2.3 + 3.x.32 + 33 = 0
(x + 3)3 = 0
x + 3 = 0
x = 0 - 3
x = -3