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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)
a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
Có lẽ bạn ghi sai đề rồi nha.
a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Bài 2:
a)A= \(6x^2\)\(-11x+3\)
<=>A=\(6x^2\)\(-2x-9x+3\)
<=>A=(\(6x^2\)\(-2x\))-\(\left(9x-3\right)\)
=>A=\(2x\left(3x-1\right)\)\(-3\left(3x+1\right)\)
<=>A=\(2x\left(3x-1\right)+3\left(3x-1\right)\)
=>A=(3x-1)(2x+3)
a, \(x^2+2xy+y^2-x-y-12\)
\(=\left(x^2+2xy+y^2\right)-\left(x+y\right)-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)^2+3\left(x+y\right)-4\left(x+y\right)-12\)
\(=\left[\left(x+y\right)^2+3\left(x+y\right)\right]-\left[4\left(x+y\right)+12\right]\)
\(=\left(x+y\right).\left[\left(x+y\right)+3\right]-4.\left[\left(x+y\right)+3\right]\)
\(=\left[\left(x+y\right)+3\right].\left[\left(x+y\right)-4\right]\)
b,B = \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(t=x^2+x+1\Rightarrow t+1=x^2+x+2\)
\(\Rightarrow B=t.\left(t+1\right)-12\)
\(B=t^2+t-12\)
\(B=t^2-3t+4t-12\)
\(B=\left(t^2-3t\right)+\left(4t-12\right)\)
\(B=t.\left(t-3\right)+4.\left(t-3\right)=\left(t-3\right).\left(t-4\right)\)
Mà \(t=x^2+x+1\) nên
\(B=\left(x^2+x+1-3\right).\left(x^2+x+1-4\right)\)
\(B=\left(x^2+x-2\right).\left(x^2+x-3\right)\)
\(B=\left(x^2-x+2x-2\right).\left(x^2+x-3\right)\)
\(B=\left[\left(x^2-x\right)+\left(2x-2\right)\right].\left(x^2+x-3\right)\)
\(B=\left[x.\left(x-1\right)+2.\left(x-1\right)\right].\left(x^2+x-3\right)\)
\(B=\left(x-1\right).\left(x+2\right).\left(x^2+x-3\right)\)
Chúc bạn học tốt!!!
a) (x+y)2 - (x+y) -12
=(x+y)2- 4(x+y) + 3(x+y)-12
=(x+y)(x+y-4) + 3 ( x+y -4)
=(x+y-4)(x+y+3)