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\(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
\(125x^3+y^3=\left(5x+y\right)\left(25x^2-5xy+y^2\right)\)
A) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
B) \(x^3-\dfrac{1}{8}\)
\(=x^3-\left(\dfrac{1}{2}\right)^3\)
\(=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
C) \(8x^3+y^3\)
\(=\left(2x\right)^3+y^3\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
D) \(8x^3-27y^3\)
\(=\left(2x\right)^3-\left(3y\right)^3\)
\(=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
a)\(\left(x+3\right)\left(x^2-3x+9\right)\)
b)\(\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c)\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d)\(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
\(l,=5x\left(y^2-2yz+5z\right)\\ m,=\left(x+1\right)^3-27y^3\\ =\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\\ n,=\left(x-3y\right)^2\\ o,=\left(x+2y\right)^3\\ p,=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\\ q,=\left(x+2y\right)^2-2\left(x-2y\right)+1\\ =\left(x+2y-1\right)^2\)
a: \(50x^5-8x^3\)
\(=2x^3\left(25x^2-4\right)\)
\(=2x^3\left(5x-2\right)\left(5x+2\right)\)
b: \(x^4-5x^2-4y^2+10y\)
\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)
\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)
c: \(36a^2+12a+1-b^2\)
\(=\left(6a+1\right)^2-b^2\)
\(=\left(6a+1-b\right)\left(6a+1+b\right)\)
d: \(x^3+y^3-xy^2-x^2y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\cdot\left(x-y\right)^2\)
e: Ta có: \(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
f: Ta có: \(9x^4+16x^2-4\)
\(=9x^4+18x^2-2x^2-4\)
\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(9x^2-2\right)\)
g: Ta có: \(-6x^2+5xy+4y^2\)
\(=-6x^2+8xy-3xy+4y^2\)
\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left(-2x-y\right)\)
h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)
\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)
\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Bài 1
a) (15x4 - 8x3 + 3x2) : 3x2 = 15x4 : 3x2 - 8x3 : 3x2 + 3x2 : 3x2 = 5x2 - \(\dfrac{8}{3}x\) + 1
b) (4x2 - 4xy + y2) : (2x - y) = (2x - y)2 : (2x - y) = 2x - y
c) dùng phương pháp chia đa thức 1 biến đã sắp xếp nha
d) (x2 - 2x + 1) : (x - 1) = (x - 1)2 : (x - 1) = x - 1
Bài 2
a) x2 + 3x + 3xy + 9xy = x2 + 3x + 12xy = x (x + 3 + 4y)
b) x2 - y2 + 2x + 1 = x2 + 2x + 1 - y2 = (x + 1)2 - y2 = (x + 1 - y)(x + 1 + y)
c) x2 - xy + x - y = x (x - y) + (x - y) = (x - y)(x + 1)
x4,x2,y2 là mũ nhá các bn
giúp mk nhanh lên mình sắp phải nộp r
a) \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b) \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c) \(125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
a, \(M=x^2-100=\left(x-10\right)\left(x+10\right)\)
b, \(N=x^3+1000=\left(x+10\right)\left(x^2-10x+100\right)\)
c, \(P=125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
d, \(Q=8x^3-27=\left(2x-3\right)\left(2x+3\right)\)