Phân tích đa thức thành nhân tử
9x^2-4y^2+16xy
(3m-1)^2-4m
21-(2k+5)(k+3)
(4p+5)(p-2)+(p+2)^2
(y+7)(3y-4)+(1-y)(5y+2)-6
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Bạn tách 3 - 4 câu thành 1 phần câu hỏi rồi gửi chứ dài quá nhiều người ngại trả lời lắm :(
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
2:
a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)
\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)
b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-1\right)\)
c: \(=\left(y^2+10y+25\right)-9z^2\)
\(=\left(y+5\right)^2-\left(3z\right)^2\)
\(=\left(y+5+3z\right)\left(y+5-3z\right)\)
d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)
\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)
1:
a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)
b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)
\(=2y\left(5y-6\right)+4\left(5y-6\right)\)
\(=2\left(5y-6\right)\left(y+2\right)\)
c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)
\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)
\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)
d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)
\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)
\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)
\(=2y\left(x+y\right)\left(-x-7y\right)\)
Bài 1
a) x(3 - 4x) + 5(3 - 4x)
= (3 - 4x)(x + 5)
b) 2y(5y - 6) - 4(6- 5y)
= 2y(5y - 6) + 4(5y - 6)
= (5y - 6)(2y + 4)
= 2(5y - 6)(y + 2)
c) 27(x - 2)³ - 3x(2 - x)²
= 27(x - 2)³ - 3x(x - 2)²
= 3(x - 2)²[9(x - 2) - x]
= 3(x - 2)²(9x - 18 - x)
= 3(x - 2)²(8x - 18)
= 6(x - 2)²(4x - 9)
d) 6y(x² - y²) - 8y(x + y)²
= 6y(x - y)(x + y) - 8y(x + y)²
= 2y(x + y)[3(x - y) - 4(x + y)]
= 2y(x + y)(3x - 3y - 4x - 4y)
= 2y(x + y)(-x - 7y)
= -2y(x + y)(x + 7y)
a) 3x+9y
=3(x+3y)
b) 4x2-16xy
=4x(x-4y)
c) x2-6x+9
=(x-3)2
d) 4y2-36
=(2y-6)(2y+6)
e) 4y2-4
=(2y-2)(2y+2)
f) 5a2-9
=\(\left(\sqrt{5}a-3\right)\left(\sqrt{5}a+3\right)\)
g) x3-2x2+x
=x(x2-2x+1)
=x(x-1)2
h) 5xy3-10xy2+5xy
=5xy(y2-2y+1)
=5xy(y-1)2
i) 12x2y+12xy+3y
=3y(4x2+4x+1)
=3y(2x+1)2
k) 5x-5y+y2-xy
=(5x-5y)+(y2-xy)
=5(x-y)-y(x-y)
=(x-y)(5-y)
#H
Trả lời:
5, x2 - y2 + 4x + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y ) ( x + 2 + y )
6, x2 + 2x - 4y2 - 4y
= ( x2 - 4y2 ) + ( 2x - 4y )
= ( x - 2y ) ( x + 2y ) + 2 ( x - 2y )
= ( x - 2y ) ( x + 2y + 2 )
7, 3x2 - 4y + 4x - 3y2
= ( 3x2 - 3y2 ) + ( 4x - 4y )
= 3 ( x2 - y2 ) + 4 ( x - y )
= 3 ( x - y ) ( x + y ) + 4 ( x - y )
= ( x - y ) [ 3 ( x + y ) + 4 ]
= ( x - y ) ( 3x + 3y + 4 )
8, x4 - 6x3 + 54x - 81
= ( x4 - 81 ) - ( 6x3 - 54x )
= ( x2 - 9 ) ( x2 + 9 ) - 6x ( x2 - 9 )
= ( x2 - 9 ) ( x2 + 9 - 6x )
= ( x - 3 ) ( x + 3 ) ( x - 3 )2
= ( x - 3 )3 ( x + 3 )
a, \(x^2-y^2+4x+4=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
b, \(x^2+2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)+2\left(x-2y\right)=\left(x-2y\right)\left(x+2+2y\right)\)
c, \(3x^2-4y+4x-3y^2=3\left(x-y\right)\left(x+y\right)-4\left(y-x\right)=\left(x-y\right)\left(3x+3y+4\right)\)
d, \(x^4-6x^3+54x-81=\left(x^2+9\right)\left(x-3\right)\left(x+3\right)-6x\left(x^2-9\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\left(x+3\right)\)
a) \(39x-39y=39\left(x-y\right)\)
b) \(3x^2\left(x-3y\right)-5y\left(3y-x\right)=3x^2\left(x-3y\right)+5y\left(x-3y\right)\)
\(=\left(3x^2+5x\right)\left(x-3y\right)=x\left(3x+5\right)\left(x-3y\right)\)
c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)
d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)
e) \(7x^2-7xy+5x-5y=7x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(7x+5\right)\)
f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)
\(=5\left[x^2-\left(3y+1\right)^2\right]=5\left(x-3y-1\right)\left(x+3y+1\right)\)
g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v )
\(=\left(x+1\right)^2-\left(y+1\right)^2=\left(x+1-y-1\right)\left(x+1+y+1\right)=\left(x-y\right)\left(x+y+2\right)\)
h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x+5\right)\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Đề sai rồi bạn