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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)
a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
\(\text{a) }\left(x^2+x\right)^2-2\left(x^2+x\right)-15\\ \text{Đặt }\left(x^2+x\right)^2=y\\ \text{Ta được: }\left(x^2+x\right)^2-2\left(x^2+x\right)-15\\ \\ =y^2-2y-15\\ \\ =y^2+3y-5y-15\\ =\left(y^2+3y\right)-\left(5y+15\right)\\ \\ =y\left(y+3\right)-5\left(y+3\right)\\ \\ =\left(y+3\right)\left(y-5\right)\\ Thay\text{ }y=x^2+x\text{ vào đa thức }\\ Ta\text{ lại được: }\left(y+3\right)\left(y-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
\(\text{b) }\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\\ \left(x^2+3x+1\right)\left(x^2+3x+1+1\right)-6=\\ \text{Đặt }\left(x^2+3x+1\right)=y\\ \text{Ta được: }\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\\ =y\left(y+1\right)-6\\ \\ =y^2+y-6\\ \\ =y^2+3y-2y-6\\ \\ =\left(y^2+3y\right)-\left(2y+6\right)\\ \\ =y\left(y+3\right)-2\left(y+3\right)\\ \\ =\left(y-2\right)\left(y+3\right)\\ Thay\text{ }y=\left(x^2+3x+1\right)\text{ vào đa thức }\\ \text{Ta lại được: }\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\\ \\ =\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(\text{c) }\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)\\ =\left(x^2+x+7x+7\right)\left(x+3\right)\left(x+5\right)\\ =\left[\left(x^2+x\right)+\left(7x+7\right)\right]\left(x+3\right)\left(x+5\right)\\ =\left[x\left(x+1\right)+7\left(x+1\right)\right]\left(x+3\right)\left(x+5\right)\\ \\=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)\)
( x2 + 8x + 7 ) ( x + 3 )( x +5)
= ( x2 + 2.4x + 42 - 9 )( x + 3)( x + 5)
= [( x + 4)2 - 32] .( x + 3)(x + 5)
= ( x +4 +3)( x + 4 - 3)( x + 3)( x + 5)
= ( x + 7)( x + 1)( x + 3)( x + 5)
Đặt : x + 4 = y , ta có :
( y + 3)( y - 3)( y - 1)(y + 1)
=( y2 - 32)( y2 - 1)
= y4 - y2 - 9y2 + 9
= y2( y2 - 1) - 9( y2 - 1)
= ( y + 3)( y - 3)( y - 1)( y + 1)
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