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(1) đa thức A\(⋮̸\) B vì \(7x⋮̸\)3x2
(2) đa thức A\(⋮̸\) B vì 2ab3c2 \(⋮̸\) -5a2bc2
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
1, 5a2xy-10a3x-15ay = 5a( axy - 2a\(^2\)x - 3y )
2, mxy-m2x+my = m( xy - mx + y )
3, 2mx-4m2xy+6mx = 2mx( 1 - 2my + 3 ) = 2mx( 4 - 2my )
4, a2b-2ab2+ab = ab( a - 2b + 1 )
5, 5a2b-2ab2+ab = ab( 5a - 2b +1 )
6, -3x2y3-6x3y2-x2y2 = -3x\(^2\)y\(^2\) ( y + 2x + 1 )
7, 5x2y4-10x4y2+5x2y2 = 5x\(^{^2y^2}\)( y\(^2\) - 2x\(^2\) + 1 )
8, -2x3y4-4x4y3+2x3y3 = 2\(x^3y^3\) ( -y - 2x + 1 )
9, 4x3y2-8x3y+16xy2-24 = 4( x\(^3\)y\(^2\) - 2x\(^3\)y + 4 xy\(^2\) - 6 )
10, 12x3y-6xy+3x = 3x( 4x\(^2\)y - 2y + 1 )
11, 2(x-y)-a(x-y) = ( 2 - a ) ( x - y )
12, a(x-y)+b(x-y)= ( a + b ) ( x - y )
13, m(x+y)-n(x+y) = ( m - n ) ( x + y )
14, 2a(x+y)-4(x+y) = ( 2a - 4 )( x + y ) = 2( a - 2 ) ( x + y )
15, 3a(x+y)-6ab(x+y) = ( 3a - 6ab )( x + y ) = 3a( 1 - 2b ) ( x + y )
16, 5a2(x-y)+10a(x-y) = ( 5a\(^2\)+10a )( x - y ) = 5a( a + 2 ) ( x - y )
17, -2ab(x-y)-4a(x-y) = ( -2ab - 4a )( x - y ) = -2a( b + 2 )( x - y )
18, 3a(x-y)+2(x-y) = ( 3a + 2 ) ( x - y )
19, m(a-b)-m2(a-b) = ( m - m\(^2\) ) ( a - b ) = m( 1 - m ) ( a - b )
20, mx(a+b)-m(a+b) = ( mx - m ) ( a + b ) = m( x - 1 )( a + b )
21, x(a-b)-y(b-a) = x( a - b ) + y( a - b ) = ( x + y ) ( a - b )
22, ab(x-5)-a2(5-x) = ab( x - 5 ) + a\(^2\)( x - 5 ) = ( ab + a\(^2\) ) ( x - 5 ) = a( b + a )( x - 5 )
23, 2a2(x-y)-4a(y-x)= 2a\(^2\)( x - y ) + 4a( x - y )=( 2a\(^2\) + 4a ) ( x - y )= 2a( a + 2 )( x - y )
Đăng ít thôi =))
a. \(5a^2xy-10a^3x-15ay=5a\left(axy-2a^2x-3y\right)\)
b. \(mxy-m^2x+my=m\left(xy-mx+y\right)\)
c. \(2mx-4m^2xy+6mx=2mx\left(1-2my+3\right)=2mx\left(-2my+4\right)\)
d. \(a^2b-2ab^2+ab=ab\left(a-2b+1\right)\)
e. \(5a^2b-2ab^2+ab=ab\left(5a-2b+1\right)\)
g.
A = 25x3y2 và B = 7xy3
=> A\(⋮̸\) B vì biến y của biểu thức A bé hơn biến y của biểu thức B
A = -3a4b5c và B = 2ab4
=> A\(⋮\)B = \(\dfrac{-3}{2}\)a3bc (vì tất cả các biến của biểu thức A đều có trong biến của biểu thức B và số mũ cùa B ko lớn hơn A)
a, \(\dfrac{3a^2b-4ab^2}{5ab}=\dfrac{ab\left(3a-4b\right)}{5ab}=\dfrac{3a-4b}{5}\)
b, \(\dfrac{3x^3y^2-5x^2y^3+4x^3y^3}{x^2y^2}=\dfrac{x^2y^2\left(3x-5y+4xy\right)}{x^2y^2}\)
\(=3x-5y+4xy\)
c, \(\dfrac{2a^5b^4+3a^4b^3}{-3a^4b^5}=\dfrac{a^4b^3\left(2ab+3\right)}{-3a^4b^5}=\dfrac{2ab+3}{-3b^2}\)
d, \(\dfrac{-a^5b^4+3a^6b^2}{4a^4b^2}=\dfrac{-a^4b^2\left(ab^2+3a^2\right)}{4a^4b^2}=\dfrac{-\left(ab^2+3a^2\right)}{4}\)
Chúc bạn học tốt!!!
a. \(\left(3a^2b-4ab^3\right):5ab=3a^2b:5ab-4ab^3:5ab=\dfrac{3}{5}a-\dfrac{4}{5}b^2\)
b. \(\left(3x^3y^2-5x^2y^3+4x^3y^3\right):x^2y^2=3x^3y^2:x^2y^2-5x^2y^3:x^2y^2+4x^3y^3:x^2y^2=3x-5y+4xy\)
c. \(\left(2a^5b^4+3a^4b^3\right):\left(-3a^4b^5\right)=2a^5b^4:\left(-3a^4b^5\right)+3a^4b^3:\left(-3a^4b^5\right)=-\dfrac{2a}{3b}-\dfrac{1}{b^2}\)
d. \(\left(-a^5b^4+3a^6b^2\right):4a^4b^2=\left(-a^5b^4\right):4a^4b^2+3a^6b^2:4a^4b^2=-\dfrac{1ab^2}{4}+\dfrac{3a^2}{4}\)
Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(3x\left(x-a\right)+5a^2-5ax\)
\(=3x\left(x-a\right)+5a\left(a-x\right)\)
\(=3x\left(x-a\right)-5a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-5a\right)\)
b) Ta có: \(x^3+8y^3+6x^2y+12xy^2\)
\(=x^3+3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)
\(=\left(x+2y\right)^3\)
c) Ta có: \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7x+7\)
\(=3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
f) Ta có: \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=a^2b-ab^2+b^2c-bc^2+c^2a-ca^2\)
\(=abc+a^2b-ab^2+b^2c-bc^2+c^2a-ca^2-abc\)
\(=\left(a^2b-abc\right)-\left(ab^2-b^2c\right)-\left(bc^2-ac^2\right)-\left(a^2c-abc\right)\)
\(=ab\left(a-c\right)-b^2\left(a-c\right)-c^2\left(b-a\right)-ac\left(a-b\right)\)
\(=\left(a-c\right)\left(ab-b^2\right)-c^2\left(b-a\right)+ac\left(b-a\right)\)
\(=b\left(a-c\right)\left(a-b\right)-\left(b-a\right)\left(c^2-ac\right)\)
\(=b\left(a-c\right)\left(a-b\right)+\left(a-b\right)\cdot c\cdot\left(c-a\right)\)
\(=b\left(a-c\right)\left(a-b\right)-c\left(a-b\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
g) Ta có: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)
Bài 1:
a) 25x2 - 10xy + y2 = (5x - y)2
b) 81x2 - 64y2 = (9x)2 - (8y)2 = (9x - 8y)(9x + 8y)
c) 8x3 + 36x2y + 54xy2 + 27y3
= 8x3 + 27y3 + 36x2y + 54xy2
= (2x + 3y)(4x2 - 6xy + 9y2) + 18xy(2x + 3y)
= (2x + 3y)(4x2 - 6xy + 18xy + 9y2)
= (2x + 3y)(4x2 + 12xy + 9y2)
= (2x + 3y)(2x + 3y)2 = (2x + 3y)3
c) (a2 + b2 - 5)2 - 4(ab + 2)2 = (a2 + b2 - 5)2 - 22(ab + 2)2
= (a2 + b2 - 5)2 - (2ab + 4)2
= (a2 + b2 - 5 - 2ab - 4)(a2 + b2 - 5 + 2ab + 4)
= (a2 - 2ab + b2 - 9)(a2 + 2ab + b2 - 1)
= \(\left [ (a - b)^{2} - 3^{2} \right ]\)\(\left [ (a + b)^{2} - 1\right ]\)
= (a - b - 3)(a - b + 3)(a + b - 1)(a + b + 1)
pn đăng mỗi lần vài bài thôi chứ đăng nhìn ngán lắm
Bài 2:
a) 2x3 + 3x2 + 2x + 3
= 2x3 + 2x + 3x2 + 3
= 2x(x2 + 1) + 3(x2 + 1)
= (x2 + 1)(2x + 3)
b)x3z + x2yz - x2z2 - xyz2
= xz(x2 + xy - xz - yz)
= \(xz\left [ x(x + y) - z(x + y) \right ]\)
= xz(x + y)(x - z)
c) x2y + xy2 - x - y
= xy(x + y) - (x + y)
= (x + y)(xy - 1)
d) 8xy3 - 5xyz - 24y2 + 15z
= 8xy3 - 24y2 - 5xyz + 15z
= 8y2(xy - 3) - 5z(xy - 3)
= (xy - 3)(8y2 - 5z)
e) x3 + y(1 - 3x2) + x(3y2 - 1) - y3
= x3 - y3 + y - 3x2y + 3xy2 - x
= (x - y)(x2 + xy + y2) - 3xy(x - y) - (x - y)
= (x - y)(x2 + xy + y2 - 3xy - 1)
= (x - y)(x2 - 2xy + y2 - 1)
= \((x - y)\left [ (x - y)^{2} - 1 \right ]\)
= (x - y)(x - y - 1)(x - y + 1)
câu f tương tự
các bạn ơi, giúp mk vs ngày mai mk phải học rồi!!!
help me-.- help me :)