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1) \(4a\left(x-5\right)-2\left(5-x\right)\)
\(=4a\left(x-5\right)+2\left(x-5\right)\)
\(=2\left(x-5\right)\left(2a+1\right)\)
2) \(-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(=-3a\left(x-3\right)+a^2\left(x-3\right)\)
\(=a\left(x-3\right)\left(-3+a\right)\)
3) \(2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(=2a^2b\left(x+y\right)+4a^3b\left(x+y\right)\)
\(=2a^2b\left(x+y\right)\left(1+2a\right)\)
4) \(-3a\left(x-3\right)-a^2\left(3-a\right)\)
Mình nghĩ câu này đề sai và hình như nó là câu 2 thì phải
5) \(x^{m+1}-x^m\)
\(=x^m.x-x^m\)
\(=x^m\left(x-1\right)\)
6) \(x^{m+1}+x^m\)
\(=x^m.x+x^m\)
\(=x^m\left(x+1\right)\)
7) \(x^{m+2}-x^m\)
\(=x^m.x^2-x^m\)
\(=x^m\left(x^2-1\right)\)
\(=x^m\left(x-1\right)\left(x+1\right)\)
8) \(x^{m+2}-x^2\)
\(=x^m.x^2-x^2\)
\(=x^2\left(x^m-1\right)\)
9) \(x^{m+2}-x^{m+1}\)
\(=x^{m+1}.x-x^{m+1}\)
\(=x^{m+1}\left(x-1\right)\)
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
\(1.4a\left(x-5\right)-2\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(4a+2\right)\)
\(\Leftrightarrow2\left(x-5\right)\left(2a+1\right)\)
\(2.-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(\Leftrightarrow\left(x-3\right)\left(a^2-3a\right)\)
\(\Leftrightarrow a\left(x-3\right)\left(a-3\right)\)
\(3.\text{ }\text{ }\text{ }2a^2b.\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(2a^2b+4a^3b\right)\)
\(\Leftrightarrow2a^2b\left(x+y\right)\left(1+2a\right)\)
1, 4a(x-5) +2(x-5) 2, 3a(3-x)-a^2(3-x) 3, 2a^2 b(x+y)+4a^3 b(x+y)
(4a+2)(x-5) (3-a)a(3-x) 2a^2 b (1+2a)(x+y)
minh khong lam het duoc vi khong co thoi gian
1 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
2 ) \(a^2\left(x+y\right)-b^2\left(x+y\right)\)
\(=\left(a^2-b^2\right)\left(x+y\right)\)
\(=\left[\left(a-b\right).\left(a+3\right)\right]\left(x+y\right)\)
3 ) \(6a^2-3a+12ab\)
\(=3a.2a-3a+3a.4b\)
\(=3a.\left(2a-1+4b\right)\)
4 ) \(2x^2y^4-2x^4y^2+6x^3y^3\)
\(=2x^2y^2.y^2-2x^2y^2.x^2+2x^2y^2.3xy\)
\(=2x^2y^2\left(y^2-x^2+3xy\right)\)
5 ) \(\left(x+y\right)^3-x\left(x+y\right)^2\)
\(=\left(x+y\right)^2.\left(x+y-x\right)\)
\(=\left(x+y\right)^2.y\)
1)a(m+n)+b(m+n)
=(a+b)(m+n)
2)a2(x+y)-b2(x+y)
=(a2-b2)(x+y)
3)6a2-3a+12ab
=3a.2a-3a.(1-4b)
=3a.(2a-1+4b)
5)(x+y)3-x(x+y)2
=(x+y)(x+y)2-x(x+y)2
=(x+y)2(x+y-x)
1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)
a , \(-q^3+12q^2x-48qx^2+64x^3\)
\(=-\left(q^3-12q^2x+48qx^2-64x^3\right)\)
\(=\)\(-\left(q-4x\right)^3\)
b , x2 + 2xy - y2 - 9
= - ( x2 - 2xy + y2 ) - 9
= - ( x - y )2 - 9
= ( - x + y - 3 ) ( x - y + 3 )
3 , 1 - m2 + 2mn - n2
= 1 - ( m2 - 2mn + n2 )
= 1 - ( m - n )2
= ( 1 - m + n ) ( 1 + m - n )
4 , x3 - 8 + 6a2 - 12a
= x3 + 6a2 - 12a + 8
= x3 + 6a2 - 12a + 4 + 4
= x3 + ( 6a2 - 12a + 4 ) + 4
= x3 + ( 3a - 2 )2 + 4
= ( x + 3a - 2 + 2 ) ( x2 + 3a + 2 + 2 )
( Mai làm tiếp mấy ý sau '-' muộn rồi ~ )
5 , x2 - 2xy + y2 - xz - yz
= ( x2 - 2xy + y2 ) - ( xz + yz )
= ( x - y )2 - z ( x + y )
= ( x - y ) 2 - z ( x - y )
= ( x - y ) ( x - y - z )
6 , x2 - 4xy + 4y 2 - z2 + 4z - 4t2
=( x2 - 4xy + 4y 2 ) - (z2 - 4z +4 ) . t2
= ( x - y )2 - ( z - 2 )2 . t2
= ( x - y - z - 2 ) ( x - y + z - 2 ) t2
7 , 25 - 4x2 - 4xy - y2
= 25 + ( - 4x2 - 4xy + y2 )
= 25 + ( 2x - y )2
= ( 5 + 2x - y ) ( 5 + 2x + y )
8 ,
x3 + y3 + z3 - 3xyz
= (x+y)3 - 3xy (x - y ) + z3 - 3xyz
= [ ( x + y)3 + z3 ] - 3xy ( x + y + z )
= ( x + y + z )3 - 3z ( x + y )( x + y + z ) - 3xy ( x - y - z )
= ( x + y + z )[( x + y + z )2 - 3z ( x + y ) - 3xy ]
= ( x + y + z )( x2 + y2 + z2 + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= ( x + y + z)(x2 + y2 + z2 - xy - xz - yz)
1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)
2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)
\(=25\left(a-b\right)^2=25\cdot100=2500\)
1: =(4x-1)^2-3(4x-1)
=(4x-1)(4x-1-3)
=4(x-1)(4x-1)
2: =-8x^4y^5(2y+3x)
3: =(a-5)^2-4b^2
=(a-5-2b)(a-5+2b)
5: =x^2-mx-nx+mn
=x(x-m)-n(x-m)
=(x-m)(x-n)
6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)
=(-6a-18)(8a^2-18)
=-6(2a-3)(2x+3)(a+3)
1/ 2a + 2b = 2( a + b )
2/ 3a - 6b - 9c = 3( a - 2b - 3c )
3/ 5ax - 15ay + 20a = 5a( x - 3y + 4 )
4/ 3a2x - 6a2y + 12a = 3a( ax - 2ay + 4 )
5/ 4a( x - 5 ) - 2( 5 - x ) = 4a( x - 5 ) + 2( x - 5 ) = ( x - 5 )( 4a + 2 ) = ( x - 5 )2( 2a + 1 )
6. -3a( x - 3 ) + ( 3 - x ) = 3a( 3 - x ) + 1( 3 - x ) = ( 3a + 1 )( 3 - x )
7/ xm+1 - xm = xm( x + 1 )
8/ xm+2 - x2 = x2( xm - 1 )