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
Bài 1:Phân tích đa thức thành nhân tử
1)8x^3-5xyz-24y^2+15z
2)x^4-x^3-x+1
3)25x^2(x-y)-x+y
4)16x62(z^2-y^2)z^2+y^2
5)x^3+x^2y-x^2z-xyz
6)12x^5y+24x^4y^2+12x^3y^3
7)x^9+x^8-x-1
9)x^2+7x+12
10)3x^2-8x+5
11)x^2-5xy=6y^2
câu cuối mình ghi sai xíu:x62-5xy+6y^2
Bạn tách 3 - 4 câu thành 1 phần câu hỏi rồi gửi chứ dài quá nhiều người ngại trả lời lắm :(
câu 1 và câu cuối mk ghi xíu:8xy^3-5xyz-24y^2+15z và câu câu cuối là x^2-5xy+6y^2
Phân tích các đa thức sau đây thành nhân tử
a, 36x^2 - ( 3x -2 ) ^2
b, 16(4x+5)^5 - 25 (2x+2)^2
c, ( x - y + 4 )^2
d, (x+1)^4 - (x-1)^4
e, 16x^2 - 24xy + 9y^2
f, -x^4/4 + 2x^2y^3 - 4y^6
g , 64x^3 +1
h, x^3y^6z^9 - 125
k, 27x^6 - 8x^3
I , x^6 - y^6
m, 27x^3 - 54x^2y + 36xy^2 - 8y^3
n, y^9 - 9x^2y^6 + 27x^4y^3 - 27x^6
làm ơn giải chi tiết giúp mik vs ạ , cảm ơn
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)
bài 1: phân tích các đa thức thành nhân tử:
a)x(3-4x)+5(3-4x)
b)2y(5y-6)-4(6-5y)
c)27(x-2)^3-3x(2-x)^2
d)6y(x^2-y^2)-8y(x+y)^2
bài 2:
a)2x^2-xy+2xz-yz
b)x^2-x+2y-4y^2
c)y^2+10y-9z^2+25
d) (x+2y)^3-x^2+4y^2
giúp mik nhé mik cần gấp ạ
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)
phân tích đa thức thành nhân tử : a ,4y2 + 4 + y2 b, ( 3y + 9 )3 c , ( 6 + 3y )5
Phân tích đa thức thành nhân tử:
1,3x+9y
2,4x^2-16xy
3,x^2-6x+9
4,4y^2-36
5,4y^2-4
6,5a^2-9
7,x^3-2x^2+x
8,5xy^3-10xy^2+5xy
9,12x^2y+12xy+3y
10,5x-5y+y^2-xy
Phân tích đa thức thành nhân tử:
1,3x+9y
2,4x^2-16xy
3,x^2-6x+9
4,4y^2-36
5,4y^2-4
6,5a^2-9
7,x^3-2x^2+x
8,5xy^3-10xy^2+5xy
9,12x^2y+12xy+3y
10,5x-5y+y^2-xy
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
bài 48; phân tích các đa thức sau thành nhân tử
5, x mũ 2 - y mũ 2 + 4x + 4
6, x mũ 2 + 2x - 4y mũ 2 - 4y
7, 3x mũ 2 - 4y + 4x - 3y mũ 2
8, x mũ 4 - 6x mũ 3 + 54x - 81
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, \(x^4-6x^3+54x-81=\left(x^2+9\right)\left(x-3\right)\left(x+3\right)-6x\left(x^2-9\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\left(x+3\right)\)
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)(x-3)3
#H
phân tích các đa thức sau thành nhân tử:
a)39x-39y
b)3x^2.(x-3y)-5y.(3y-x)
c)16x^2+24xy+9y^2
d)25x^2-1/25y^2
e)7x^2-7xy+5x-5y
f)5x^2-45y^2-30y-5
g)x^2+2x+1-y^2+4y-1
h)4x^2+8x-5
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 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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)