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1. 3x4 + 6x3 - 33x2 - 24x + 48
Đặt t=x2-4
=>3x4 + 6x3 - 33x2 - 24x + 48
=3.(x4+2x3-11x2-8x+16)
=3.(x4-8x2+16+2x3-8x-3x2)
=3.[(x4-8x2+16)+(2x3-8x)-3x2]
=3.[(x2-4)2+2x.(x2-4)-3x2]
thay t=x2-4 ta được :
3.(t2+2xt-3x2)
=3.(t2+2xt-3x2)
=3.(t2-xt+3xt-3x2)
=3.[t.(t-x)+3x.(t-x)]
=3.(t-x)(t+3x)
thay t=x2-4 ta được:
3.(x2-4-x)(x2-4+3x)
=3.(x2-x-4)(x2-x+4x-4)
=3.(x2-x-4)[x.(x-1)+4.(x-1)]
=3.(x2-x-4)(x-1)(x+4)
Câu còn lại tương tự
câu trên tui viết nhầm: viết lại:
x^3+3x^2+3x+2
= (3x^2+3x)+x^3+2
= 3x(x+1)+(x+1)+(x+1)+x
= (x+1)(3x+1+1)+x
a) x2 - 4x + 2 = (x2 - 4x + 4) - 2 = (x - 2)2 - 2 = \(\left(x-2+\sqrt{2}\right)\left(x-2-\sqrt{2}\right)\)
b) x2 - 12x + 11 = x2 - x - 11x + 11 = x(x - 1) - 11(x - 1) = (x - 1)(x - 11)
c) 3x2 + 6x - 9 = 3x2 - 3x + 9x - 9 = 3x(x - 1) + 9(x - 1) = (3x + 9)(x - 1) = 3(x + 3)(x - 1)
d) 2x2 - 6x + 2 = 2(x2 - 3x + 1) = 2(x2 - 3x + 9/4 - 5/4) = 2[(x - 3/2)2 - 5/4] = \(2\left(x-\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\left(x-\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\)
1.
a) \(x^2-4x+2=\left(x^2-4x+4\right)-2=\left(x-2\right)^2-2=\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)\)
b) \(x^2-12x+11=\left(x^2-12x+36\right)-25=\left(x-6\right)^2-5^2=\left(x-6-5\right)\left(x-6+5\right)=\left(x-11\right)\left(x-1\right)\)
c) \(3x^2+6x-9=3\left(x^2+2x-3\right)=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2-6x+2=2\left(x^2-3x+1\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]\)
\(=2\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)
\(x^4+5x^3-12x^2+5x+1\)
\(=x^4-x^3+6x^3-6x^2-6x^2+6x-x+1\)
\(=x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+6x^2-6x-1\right)\)
\(=\left(x-1\right)\left(x^3-x^2+7x^2-7x+x-1\right)\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)+7x\left(x-1\right)+\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x^2+7x+1\right)\)