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(x2 + 2.x.3 + 32 - 1).(x2 + 2.x.4 + 16 - 1) - 24
=[(x+3)2 - 1]. [(x+4)2-1] -24
=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24
=(x+4)(x+2)(x+5)(x-3) - 24
(x2+6x+8)(x2+8x+15)-24
<=>(x2+4x+2x+8)(x2+5x+3x+15)-24
<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24
<=> (x+4)(x+2)(x+5)(x+3)-24
<=> (x+4)(x+3)(x+2)(x+5)-24
<=>(x2+7x+12)(x2+7x+10)
đặt t=x2+7x+11 ta có:
(t-1)(t+1)-24
<=> t2-1-24
<=>t2-25
<=>(t-5)(t+5)
thay t=x2+7x+11 vào ta có:
(x2+7x+11-5)(x2+7x+11+5)
<=>(x2+7x+6)(x2+7x+16)
\(3x^4-48\)
\(=\left(3x^4-6x^3\right)+\left(6x^3-12x^2\right)+\left(12x^2-24x\right)+\left(24x-48\right)\)
\(=3x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+24\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(3x^3+6x^2\right)+\left(12x+24\right)\right]\)
\(=\left(x-2\right)\left[3x^2\left(x+2\right)+12\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(3x^2+12\right)\)
\(x^4-8x\)
\(=x\left(x^3-8\right)\)
\(=x\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(4x-8\right)\right]\)
\(=x\left[x^2\left(x-2\right)+2x\left(x-2\right)+4\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
\(b,6x-6y-x^2+xy\)
\(=\left(6x-6y\right)-\left(x^2-xy\right)\)
\(=6\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(6-x\right)\)
Bài này ko thể phân tích theo kiểu lớp 8 được (chưa học căn thức)
\(2x^2-6x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{3\sqrt{2}}{2}+\left(\frac{3\sqrt{2}}{2}\right)^2-\frac{7}{2}\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}\right)^2-\left(\frac{\sqrt{14}}{2}\right)^2\)
\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}+\frac{\sqrt{14}}{2}\right)\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}-\frac{\sqrt{14}}{2}\right)\)
\(=\left(\sqrt{2}x+\frac{\sqrt{14}-3\sqrt{2}}{2}\right)\left(\sqrt{2}x-\frac{\sqrt{14}+3\sqrt{2}}{2}\right)\)
\(2x^2-6x+1=2\left(x^2-3x+\frac{9}{4}-\frac{7}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{7}}{2}\right)^2\right]=2\left(x-\frac{3}{2}-\frac{\sqrt{7}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{7}}{2}\right)\)
\(=2\left(x-\frac{3+\sqrt{7}}{2}\right)\left(x-\frac{3-\sqrt{7}}{2}\right)\)
8x^2 + 6x - 35
= 2 × (4x^2 + 3x - 35/2)
= 2 × [(2x)^2 + 2 × 2x × 3/4 +(3/4)^2 - (9/16+35/2) ]
= 2 × [(2x+3/4)^2 -(289/16)
a)8x^2+6x-35=8x^2+20x-14x-35=4x(2x+5)-7(2x+5)=(4x-7)(2x+5)
b)2x^2-13x+20=2x^2-5x-8x+20=x(2x-5)-4(2x-5)=(x-4)(2x-5)
c)3x^2+13xy-30y^2=3x^2+18xy-5xy-30y^2=(3x-5y)(x+6y)