Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
1) \(x^3+2x-3\)
\(=\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)\)
\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+3\right)\)
2) \(x^3-6x+4\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(2x-4\right)\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x-2\right)\)
3) \(x^3-2x^2+1\)
\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)
\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x-1\right)\)
4) \(x^3+5x^2-12\)
\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)
\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+3x-6\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) x3 -2x2 +5x-4
=x3-x2-x2+x+4x-4
=x2(x-1)-x(x-1)+4(x-1)
=(x2-x+4)(x-1)
b) x3-x2+x+3
=x3+x2-2x2-2x+3x+3
=x2(x+1) -2x(x+1)+3(x+1)
=(x2-2x+3)(x+1)
c) 6x3+x2+x+1
=6x3+ 3x2-2x2-x+2x+1
=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))
=(6x2-2x+2) (x+\(\frac{1}{2}\))
=2( 3x2-x+1) (x+\(\frac{1}{2}\))
d) 4x3 + 6x2+4x+1
= 4x3+2x2+4x2+2x+2x+1
= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))
= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))
e) x6 -9x3+8
\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)
\(=\left\{\left(x+1\right)\left(x+6\right)\right\}.\left\{\left(x+3\right)\left(x+4\right)\right\}-7\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\) \(\left(1\right)\)
đặt \(x^2+7x+9=a\)
<=> \(\left(1\right)=\left(a-3\right)\left(a+3\right)-7\)
\(=a^2-16\)
\(=\left(a-4\right)\left(a+4\right)\)
hay\(\left(1\right)=\) \(\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)
\(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)
những câu còn lại cũng nhóm đầu với cuối , hai cái giữa với nhau , xong làm tương tự câu trên
học tốt
a) (x + 1)(x + 3)(x + 4)(x + 6) - 7
= (x + 1)(x + 6) (x + 3)(x + 4) - 7
= (x2 + 7x + 6)(x + 7x + 12) - 7
Đặt t = x2 + 7x + 6
Ta có : t(t + 6) - 7
= t2 + 6t - 7
= t2 + 6t + 9 - 16
= (t + 3) - 16
= (t + 3 - 4)(t + 3 + 4)
= (t - 1)(t + 7)
Nên :
Pt = (x2 + 7x + 6 - 1)(x2 + 7x + 6 + 7)
= (x2 + 7x + 5)(x2 + 7x + 13)
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
a) \(A=\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+6\right)-10\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+15\right)-10\)
Đặt \(x^2+8x+12=t\)
Khi đó ta có:
\(A=t\left(t+3\right)-10\)
\(=t^2+3t-10\)
\(=\left(t-2\right)\left(t+5\right)\)
Thay trở lại ta có:
\(A=\left(x^2+8x+10\right)\left(x^2+8x+17\right)\)
b) \(B=x\left(2x+1\right)\left(2x+3\right)\left(4x+8\right)-18\)
\(=\left(4x^2+8x\right)\left(4x^2+8x+3\right)-18\)
Đặt \(4x^2+8x=t\)
Khi đó ta có:
\(B=t\left(t+3\right)-18=t^2+3t-18=\left(t-3\right)\left(t+6\right)\)
Thay trở lại ta có:
\(B=\left(4x^2+8x-3\right)\left(4x^2+8x+6\right)=2\left(4x^2+8x-3\right)\left(2x^2+4x+3\right)\)