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a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
(x+1)(x+3)(x+5)(x+8)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
=>x2+8x+15=t+8
=>(x2 +8x+7)(x2+8x+15)+15
=t(t+8)+15
=t2+8t+15
=t2+3t+5t+15
=t(t+3)+5(t+3)
=(t+3)(t+5)
=(x2+8x+10)(x2+8x+12)
Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(\Rightarrow A=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)
\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t+1\right)\left(t-1\right)\)
\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+2x+6x+12\right)\left(x^2+8x+10\right)\)\(=\left[x\left(x+2\right)+6\left(x+2\right)\right]\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
(x + 1)(x + 2)(x + 3)(x + 4) - 24
= x4 + 10x3 + 35x2 + 50x + 24 - 24
= x4 + 10x3 + 35x2 + 50x
( x + 1 ). ( x + 2 ) ( x + 3 ) ( x + 4 ) - 24
= ( x2 + 5x + 4 ) .( x2 + 5x + 6 ) - 24
Đặt t = x2 + 5x + 5
=> ( t - 1 ). ( t + 1 ) - 24
= t2 - 1 - 24
= t2 - 25
= ( t - 5 ). ( t + 5 )
= ( x2 + 5x + 5 - 5 ) . ( x2 + 5x + 5 + 5 )
= ( x2 + 5x ) . ( x2 + 5x + 10 )
= x. ( x + 5 ) . ( x2 + 5x + 10 )
a, 4y(x-1)-(1-x)
=(x-1)(4y+1)
b,3x(z+2)+5(-x-2)
=3x(z+2)-5(x+2)
=(z+2)(3x-5)
c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1,5=a\)
\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)
\(\Rightarrow A=a^2-0,25-6\)
\(\Rightarrow A=a^2-\frac{25}{4}\)
\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)
Thay \(a=x^2+3x+0,5\)vào A ta có :
\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)
\(A=\left(x^2+3x-2\right)\left(x^2+3x+3\right)\)
c, Đặt \(x^2+3x+2=a\)
Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)
\(=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a+2\right)\left(a-3\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Câu d làm tương tự .
Gợi ý : (x+3)(x+5) = x2 + 8x + 15
đặt bằng a rồi giải tiếp