a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
Bài giải:
a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
\(a.\)\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\)
\(b.\) \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+4\right)\left(x+1\right)\)
\(c.\) \(x^2-x-6\)
\(=x^2+2x-3x-6\)
\(=x\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x-3\right)\left(x+2\right)\)
\(d.\) \(x^4+4\)
\(=x^4+4+4x^2-4x^2\)
\(=\left(x^4+4x+4\right)-4x^2\)
\(=\left(x^2+4\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
a) \(x^2-4x+3\)
\(=\left(x\right)^2-2\left(x\right).\left(2\right)+\left(2\right)^2-4+3\)
\(=\left(x-2\right)^2-1\)
\(=\left(x-2-1\right).\left(x-2+1\right)\)
\(=\left(x-3\right).\left(x-1\right)\)
b) \(x^2+5x+4\)
\(=\left(x\right)^2+2.\left(x\right).\left(\dfrac{5}{2}\right)+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+4\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{25}{4}+\dfrac{16}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\)
\(=\left(x+\dfrac{5}{2}-\dfrac{3}{2}\right).\left(x+\dfrac{5}{2}+\dfrac{3}{2}\right)\)
\(=\left(x+1\right).\left(x+4\right)\)
c) \(x^2-x-6\)
\(=\left(x\right)^2-2.\left(x\right).\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-6\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}-\dfrac{24}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}-\dfrac{5}{2}\right).\left(x-\dfrac{1}{2}+\dfrac{5}{2}\right)\)
\(=\left(x-3\right).\left(x+2\right)\)
\(a,x^2-4x+3\\ =\left(x^2-2.2x+2^2\right)-1\\ =\left(x-2\right)^2-1\\ =\left(x-2-1\right)\left(x-2+1\right)\\ =\left(x-3\right)\left(x-1\right)\)
\(b,x^2+5x+4\\ =x^2+4x+x+4\\ =\left(x^2+x\right)+\left(4x+4\right)\\ x\left(x+1\right)+4\left(x+1\right)\\ =\left(x+4\right)\left(x+1\right)\)
\(c,x^2-x-6=\\x^2+2x-3x-6\\ =\left(x^2+2x\right)-\left(3x+6\right)\\ =x\left(x+2\right)-3\left(x+2\right)\\ =\left(x-3\right)\left(x+2\right)\)
\(d,x^4+4\\ =\left(x^4+4+4x^2\right)-4x^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ \left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
nhờ mn là giúp mình với ạ , minh đang cần gấp :( 
