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a) x4 + 4 = (x4 + 4x2 + 4) - 4x2 = (x2 + 2)2 - 4x2 = (x2 + 2x + 2)(x2 - 2x + 2)
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24 = (x + 2)(x + 5)(x + 3)(x + 4) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = y => y(y + 2) - 24 = y2 + 2y - 24
= y2 + 6y - 4y - 24 = (y - 4)(y + 6) = (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16) = (x2 + x + 6x + 6)(x2 + 7x + 16) = (x + 1)(x + 6)(x2 + 7x + 16)
a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)
cho \(\left(x^2-x\right)=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=\left(a^2+6a\right)-\left(2a+12\right)\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2-x+6\right)\left(x^2-x-2\right)\)
b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Gọi \(x^2+5x+5=a\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(\Rightarrow\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
1) (3x+4)(x+1) = 3x2+7x+4 đặt là a
(6x+7)2= 36x2+84x+49 = 12a+1
=> a(12a+1)- 6 = 12a2 -a -6 = (3a+2)(4a-3) = (9x2+21x+14)(12x2+28x+13)
2) (x-2)2=x2-4x+4 đặt là a
(2x-5)(2x-3)= 4x2-16x+15 =4a-1
=> a(4a-1)-5 = 4a2-a-5 = (4a-5)(a+1) = ( 4x2-16x+11)(x2-4x+5)
3) đặt (x+3)2 =a ta làm tương tự
4) (x-2)(x-10)(x-4)(x-5) = (x2-12x+20)(x2-9x+20)
đặt x2+20=a => (a-12x)(a-9x)-54x2 = a2-21ax+54x2 = (a-18x)(a-3x) = (x2-18x+20)(x2-3x+20)
\(x^5-3x^4-x^3-x^2+3x+1\)
\(=\left(x^5-x^2\right)-\left(3x^4-3x\right)-\left(x^3-1\right)\)
\(=x^2\left(x^3-1\right)-3x\left(x^3-1\right)-\left(x^3-1\right)\)
\(=\left(x^3-1\right)\left(x^2-3x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-\frac{3}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{13}}{2}\right)\)
\(x^5-3x^4-x^3-x^2+3x+1\)\(1\)\(=\left(x^5-x^4\right)-\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)-\left(4x^2-4x\right)-\left(x-1\right)\)
\(=x^4\left(x-1\right)-2x^3\left(x-1\right)-3x^2\left(x-1\right)-4x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4-2x^3-3x^2-4x-1\right)\)
x(x+1)(x+2)(x+3) + 1
= x(x+3).(x+1)(x+2) + 1
= (x^2 + 3x) ( x^2 + 3x +2) + 1
Đặt x^2 + 3x = y ta có :
y .(y + 2)+ 1 = y^2 + 2y + 1 = (y + 1)^2
Thay y = x^2 + 3x ta có :
( y + 1)^2 = ( x^2 + 3x + 1)^2
x.(x+1).(x+2).(x+3)+1
=x.(x+3).(x+1).(x+2)+1
=(x2+3x)(x2+3x+2)+1
Đặt y=x2+3x ta được:
y.(y+2)+1
=y2+2x+1
=(y+1)2
thay y=x2+3x ta được:
(x2+3x)2
=[x.(x+3)]2
=x2.(x+3)2
Vậy x.(x+1).(x+2).(x+3)+1=x2.(x+3)2
x8 + x4 + 1
=x8+2x4+1-x4
=(x4+1)2-x4
=(x4-x2+1)(x4+x2+1)
=(x4-x2+1)(x4+2x2+1-x2)
=(x4-x2+1)[(x2+1)2-x2]
=(x4-x2+1)(x2-x+1)(x2+x+1)
x8 + x4 + 1
= ( x4 )2 + 2x4 + 1 - x4
= ( x4 + 1 )2 - x4
= ( x4 + 1 - x2 ) ( x4 + 1 + x2 )
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)
\(=\)\(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)
\(=\)\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+11=t\) ta có :
\(=\)\(\left(t-1\right)\left(t+1\right)-8\)
\(=\)\(t^2-1-8\)
\(=\)\(t^2-9\)
\(=\)\(\left(t-3\right)\left(t+3\right)\)
\(=\)\(\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)\)
\(=\)\(\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
Chúc bạn học tốt ~