Phân tích các đa thức sau thành nhân tử:
\(\left(4x-1\right)^2-121\)
\(x^6-y^6\)
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\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right).\left(x^4-x^2y^2+y^4\right)\\ ---\\ 0,04-9x^2=\left(0,2\right)^2-\left(3x\right)^2=\left(0,2-3x\right)\left(0,2+3x\right)\\ ---\\ 32x^2-2\left(y-1\right)^2=2\left[16x^2-\left(y-1\right)^2\right]=2\left[\left(4x\right)^2-\left(y-1\right)^2\right]\\ =2\left(4x-y+1\right)\left(4x+y-1\right)\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(\left(x^2+4x+6\right)\left(x^2+6x+6\right)-3x^2\left(1\right)\)
Đặt \(x^2+5x+6=t\)Thay vào (1) ta được:
\(\left(t-x\right)\left(t+x\right)-3x^2\)
\(=t^2-x^2-3x^2\)
\(=t^2-4x^2\)
\(=\left(t-2x\right)\left(t+2x\right)\)Thay \(t=x^2+5x+6\)ta được:
\(\left(x^2+5x+6-2x\right)\left(x^2+5x+6+2x\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+3x+6\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+3x+6\right)\left(x+1\right)\left(x+6\right)\)
Rút gọn thôi chứ phân tích sao được ._.
( x - 3 )2 - ( 4x + 5 )2 - 9( x + 1 )2 - 6( x - 3 )( x + 1 )
= x2 - 6x + 9 - ( 16x2 + 40x + 25 ) - 9( x2 + 2x + 1 ) - 6( x2 - 2x - 3 )
= x2 - 6x + 9 - 16x2 - 40x - 25 - 9x2 - 18x - 9 - 6x2 + 12x + 18
= -30x2 - 52x - 7
Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))
Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)
\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)
\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)
\(=\left(4x+7\right)\left(12x+17\right)\)
g ) \(4x^2\left(x-2y\right)-\left(4x+1\right)\left(2y-x\right)\)
\(=4x^2\left(x-2y\right)+\left(4x+1\right)\left(x-2y\right)\)
\(=\left(4x^2+4x+1\right)\left(x-2y\right)\)
\(=\left(2x+1\right)^2\left(x-2y\right)\)
h ) \(x^2-ax^2-y+ay+cx^2-cy\)
\(=x^2\left(1-a+c\right)-y\left(1-a+c\right)\)
\(=\left(x^2-y\right)\left(1-a+c\right)\)
\(P\left(x\right)=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right].\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right).\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right).\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x=t\), ta có:
\(\left(t+2\right)\left(t-1\right)-4\)
\(=t^2-t+2t-2-4=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
Thay \(t=12x^2+11x\), ta được:
\(P\left(x\right)=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Đs...
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
`#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).`
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: \(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)