Phân tích đa thức thành nhân tử:
a) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
b) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
Phân tích đa thức sau thành nhân tử:
a) \(x^2-2xy+3x-3y+y^2-4\)
b) \(2\left(x^2-6x+1\right)^2+5\left(x^2-6x+1\right)\left(x^2+1\right)+2\left(x^2+1\right)^2\)
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)
phân tích đa thức thành nhân tử
a, \(100x^2-\left(x^2+25\right)^2\)
b, \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
c, \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
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.
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Phân tích các đa thức sau thành nhân tử :
a/ \(10x\left(x-y\right)-6y\left(y-x\right)\)
b/ \(14x^2y-21xy^2+28x^3y^2\)
c/ \(x^2-4+\left(x-2\right)^2\)
d/ \(\left(x+1\right)^2-25\)
e/ \(x^2-4y^2-2x+4y\)
f/ \(x^2-25-2xy+y^2\)
g/ \(x^3-2x^2+x-xy^2\)
h/ \(x^3-4x^2-12x+27\)
i/ \(x^2+5x-6\)
m/ \(6x^2-7x+2\)
n/ \(4x^4+81\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Phân tích các đa thức sau thành nhân tử:
a) \({x^3} + 4x\)
b) \(6ab - 9a{b^2}\)
c) \(2a\left( {x - 1} \right) + 3b\left( {1 - x} \right)\)
d) \({\left( {x - y} \right)^2} - x\left( {y - x} \right)\)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
phân tích đa thức \(\dfrac{1}{2}x^2-2y^2\) thành nhân tử
a. \(\dfrac{1}{2}x^2-2y^2=\dfrac{1}{2}\left(x^2-4y^2\right)=\dfrac{1}{2}\left(x-2y\right)\left(x+2y\right)\)
b. \(\dfrac{1}{2}x^2-2y^2=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\)
Cách phân tích nào đúng, a hay b ?
phân tích đa thức \(\dfrac{1}{2}x^2-2y^2\) thành nhân tử
a. \(\dfrac{1}{2}x^2-2y^2=\dfrac{1}{2}\left(x^2-4y^2\right)=\dfrac{1}{2}\left(x-2y\right)\left(x+2y\right)\)
b. \(\dfrac{1}{2}x^2-2y^2=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\)
Cách phân tích nào đúng, a hay b ?
Phân tích đa thức thành nhân tử
\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+9\)
\(x^2+2xy+y^2+2x+2y-15\)
\(4x^4y^4+1\)
\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+9\)
\(=\left(x^2+5x+6\right)\left(x^2+5x\right)+9\)
Đặt \(t=x^2+5x\)ta được;
\(t\left(t+6\right)+9=t^2+6t+9\)
\(=\left(t+3\right)^2=\left(x^2+5x+3\right)^2\)
b)\(x^2+2xy+y^2+2x+2y-15\)
\(=\left(x+y+1\right)^2-4^2\)
\(=\left(x+y+1+4\right)\left(x+y+1-4\right)\)
\(=\left(x+y-3\right)\left(x+y+5\right)\)
c)\(4x^4y^4+1=\left(2x^2y^2-2xy+1\right)\left(2x^2y^2+2xy+1\right)\)
phân tích đa thức thành nhân tử :
a) \(x^2+2y^2+2xy-2x-6y+2015\)
b)\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
c) \(\left(x-y\right)^5+\left(y-z\right)^5+\left(z-x\right)^5\)