Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cái này cách tốt nhất là vào Cốc Cốc Math rồi gõ các nhân tử vào là nó sẽ ra nhé !
^_^
\(2x^4-9x^3+2x^3-9x^2+7x^2+7x+6x+6\)
\(\left(2x^4+2x^3\right)-\left(9x^3+9x^2\right)+\left(7x^2+7x\right)+\left(6x+6\right)\)
\(2x^3\left(x+1\right)-9x^2\left(x+1\right)+7x\left(x+1\right)+6\left(x+1\right)\)
\(\left(x+1\right)\left(2x^3-9x^2+7x+6\right)\)
b)\(\left(10x^4-50x^3y\right)+\left(23x^3y-115x^2y^2\right)+\left(5x^2y^2-25xy^3\right)-\left(2xy^3-10y^4\right)\)
\(10x^3\left(x-5y\right)+23x^2y\left(x-5y\right)+5xy^2\left(x-5y\right)-2y^3\left(x-5\right)\)
Bài 2:
a: \(A=\left(2x-y\right)^2=\left(12-2\right)^2=100\)
b: \(=\left(x-3\right)^3=100^3=1000000\)
c: \(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(=\left(6+4-90\right)\left(6+4+90\right)=-80\cdot100=-8000\)
Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
a) \(x\left(x-y\right)+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
b) \(2x+2y-x\left(x+y\right)\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
c) \(5x^2-5xy-10x+10y\)
\(=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)\)
d) \(4x^2+8xy-3x-6y\)
\(=4x\left(x+2y\right)-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(4x-3\right)\)
e) \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2+2y^2-2\right)-\left(x^2z-z+y^2z\right)\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2-1+y^2\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
a. gọi phần đầu đấy là A nhá, để đỡ cần viết lại
A=...............
= (3x+5)2 + ( 3x-5)2 - 9x2 -4
= (9x2 +30x + 25 ) + ( 9x2 -30x+ 25 ) - 9x2 -4
= 9x2 +30x + 25 + 9x2 -30x+25-9x2 -4
= 9x2 + 46
sai thì thôi nhé. bạn nên kiểm tra lại
d. (2x-1)*(4x2 + 2x +1 ) - 8x*( x2 +1) - 5
= 8x3 -1 - 8x3 -8x-5
= -8x-6
= -2(4x+3)
sai nhé. bạn nên kiểm tra lại
B1:
a) \(x^3-2x^2+x-2\)
= \(x^2\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+1\right)\)
b) \(2x^3+3x^2-3x-2\)
= \(2x^3-2x^2+5x^2-5x+2x-2\)
= \(2x^2\left(x-1\right)+5x\left(x-1\right)+2\left(x-1\right)\)
= \(\left(x-1\right)\left(2x^2+5x+2\right)\)
= \(\left(x-1\right)\left(2x^2+4x+x+2\right)\)
= \(\left(x-1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
c) \(5x^2+5y^2-x^2z+2xyz-y^2z-10xy\)
= \(5\left(x^2+2xy+y^2\right)+z\left(x^2+2xy+y^2\right)\)
= \(5\left(x+y\right)^2+z\left(x+y\right)^2\)
= \(\left(x+y\right)^2\left(5+z\right)\)
d) \(x^3-3x^2y+3xy^2-x+y-y^3\)
= \(\left(x-y\right)^3-\left(x-y\right)\)
= \(\left(x-y\right)\left[\left(x-y\right)^2-1\right]\)
= \(\left(x-y\right)\left(x-y-1\right)\left(x-y+1\right)\)
B2:
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\left(2x-5\right).\left(-2\right)=0\)
\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\left(x+3\right)\left(x^2-2x\right)=0\)
\(\left(x+3\right).x.\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
c) \(2x^3+3x^2+2x+3=0\)
\(x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
Ta thấy \(x^2+1>0\) với mọi x
\(\Rightarrow2x+3=0\Rightarrow x=\dfrac{-3}{2}\)
a, 5x2 - 45x = 5x(x - 9)
b, 3x3y - 6x2y - 3xy3 - 6axy2 - 3a2xy + 3xy
= 3xy(x2 - 2x - y2 - 2ay - a2 + 1)
= 3xy[ (x2 - 2x + 1) - (a2 + 2ay + y2) ]
= 3xy[ (x - 1)2 - (a + y)2 ]
= 3xy(x - 1 + a + y)(x - 1 - a - y)
f, 3xy2 - 12xy + 12x
= 3x(y2 - 4y + 4)
= 3x(y - 2)2
g, 2x2 - 8x + 8
= 2(x2 - 4x + 4)
= 2(x - 2)2
h, 5x3 + 10x2y + 5xy2
= 5x( x2 + 2xy + y2 )
= 5x(x + y)2
k, x2 + 4x - 2xy - 4y + y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
i, x3 + ax2 - 4a - 4x
= (x3 - 4x) + (ax2 - 4a)
= x(x2 - 4) + a(x2 - 4)
= (x + a)(x2 - 4)
= (x + a)(x + 2)(x - 2)
Chúc bạn học tốt !
giúp mình với
a: \(=\left(a+b+c\right)^2+2\left(a+b+c\right)\left(b-c\right)+\left(b-c\right)^2\)
\(=\left(a+b+c+b-c\right)^2=\left(a+2b\right)^2\)
b: \(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(=\left(x^4+4\right)\left(x^4-4\right)=x^8-16\)
d: \(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)-3\left(x^2+y^2+z^2\right)\)
=0