Phân tích thành nhân tử:
\(B=\left(x^2-y^2+1\right)^3-x^6+y^6-1\)
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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
13 tháng 12 2023
\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right).\left(x^4-x^2y^2+y^4\right)\\ ---\\ 0,04-9x^2=\left(0,2\right)^2-\left(3x\right)^2=\left(0,2-3x\right)\left(0,2+3x\right)\\ ---\\ 32x^2-2\left(y-1\right)^2=2\left[16x^2-\left(y-1\right)^2\right]=2\left[\left(4x\right)^2-\left(y-1\right)^2\right]\\ =2\left(4x-y+1\right)\left(4x+y-1\right)\)
NL
2
11 tháng 12 2023
`#3107.101107`
`(4x - 1)^2 - 121`
`= (4x - 1)^2 - (11)^2`
`= (4x - 1 - 11)(4x - 1 + 11)`
`= (4x - 12)(4x + 10)`
`= 4(x - 3) * 2(2x + 5)`
`= 8(x - 3)(2x + 5)`
_____
`x^6 - y^6`
`= (x^3)^2 - (y^3)^2`
`= (x^3 - y^3)(x^3 + y^3)`
`= (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)`
`= (x - y)(x + y)(x^2 + xy + y^2)`
____
Sử dụng các HĐT:
`@` `A^2 - B^2 = (A - B)(A + B)`
`@` `A^3 - B^3 = (A - B)(A^2 + AB + B^2)`
`@` `A^3 + B^3 = (A + B)(A^2 - AB + B^2).`
11 tháng 12 2023
a: \(\left(4x-1\right)^2-121\)
\(=\left(4x-1\right)^2-11^2\)
\(=\left(4x-1-11\right)\left(4x-1+11\right)\)
\(=\left(4x-12\right)\left(4x+10\right)\)
\(=8\left(x-3\right)\left(2x+5\right)\)
b: \(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
7 tháng 7 2016
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
DA
2
11 tháng 9 2018
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right).\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1-3xy\right]\)
\(=\left(x+y-1\right).\left[x^2+2xy+y^2+x+y+1-3xy\right]\)
\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
Chúc bạn học tốt.
11 tháng 9 2018
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(\left(x+y\right)^2+\left(x+y\right).1+1^2\right)-3xy\left(x+y-1\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
GH
22 tháng 8 2023
\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)
\(\left(x-y\right)^3-\left(x+y\right)^3\\ =\left(x-y-x-y\right)\left(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2\right)\\ =-2y\left(3x^2+y^2\right)\)
\(\left(x+1\right)^3+\left(x-1\right)^3\\ =\left(x+1+x-1\right)\left(x^2+2x+1-x^2+1+x^2-2x+1\right)\\ =2x\left(x^2+3\right)\)
\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)
22 tháng 8 2023
a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3
=(2x+3y)^3
b: (x-y)^3-(x+y)^3
=(x-y-x-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]
=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]
=-2y(3x^2+y^2)
c: (x+1)^3+(x-1)^3
=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]
=2x*[x^2+2x+1-x^2+1+x^2-2x+1]
=2x(x^2+3)
d: =(x-1-x-1)(x-1+x+1)
=2x*(-2)=-4x