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(3x+1)\(^2\)-(x+1)\(^2\) =(3x+1+x+1)(3x+1-x-1) = 2x(4x+2) =4x(2x+1)
Bài 1:
a,27x3+27x2+9x+1a,27x3+27x2+9x+1
=(3x)3+3.(3x)2.1+3.3x.12+13=(3x)3+3.(3x)2.1+3.3x.12+13
=(3x+1)3=(3x+1)3
b,x3+3√2x2y+6xy2+2√2y3b,x3+32x2y+6xy2+22y3
=x3+3.x2.√2y+3.x.(√2y)2+(√2y)3=x3+3.x2.2y+3.x.(2y)2+(2y)3
=(x+√2y)3=(x+2y)3
Bài 2:
a,x3+9x2+27x+27=0a,x3+9x2+27x+27=0
⇔(x+3)3=0⇔(x+3)3=0
⇔x+3=0⇔x=−3⇔x+3=0⇔x=−3
b,(x+1)3−x(x−2)2+x−1=0b,(x+1)3−x(x−2)2+x−1=0
⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0
⇔−x2+8x=0⇔−x2+8x=0
⇔−x(x−8)=0
27x³ - 27x² + 3x + 1
= 27x³ - 9x² - 18x² - 3x + 6x + 1
= (27x³ - 9x²) - (18x² - 6x) - (3x - 1)
= 9x² (3x - 1) - 6x (3x - 1) - (3x - 1)
= (3x - 1) (9x² - 6x - 1)
\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)
\(7x\left(x-5\right)+4x-20\)
\(=7x\left(x-5\right)+4\left(x-5\right)\)
\(=\left(x-5\right)\left(7x+4\right)\)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1-16\right)\left(3x-1+16\right)\)
\(=\left(3x-17\right)\left(3x+15\right)\)
\(7x\left(x-5\right)+4x-20\)
\(=7x\left(x-5\right)+4\left(x-5\right)\)
\(=\left(x-5\right)\left(7x+4\right)\)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
a. (x - 2)(x + 2) - (x - 3)2 = 9
<=> x2 - 22 - (x - 3)2 = 32
<=> x - 2 - (x - 3) = 3
<=> x - 2 - x + 3 = 3
<=> x - x = 3 - 3 + 2
<=> 0 = 2 (Vô lí)
Vậy nghiệm của PT là S = \(\varnothing\)
b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)
\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)
\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2
a) \(9x^2-12x+4=\left(3x-2\right)^2\)
b) \(25+10x+x^2=\left(x+5\right)^2\)
c) \(36x^2-25=\left(6x-5\right)\left(6x+5\right)\)
d) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\)
a: \(9x^2-12x+4=\left(3x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
c: \(36x^2-25=\left(6x-5\right)\left(6x+5\right)\)
\(x^2+3x+2=x^2+x+2x+2=x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(x+2\right)\)