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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) \(x^{m+2}-x^{m+1}\)
\(=x^{m+1}.x-x^{m+1}\)
\(=x^{m+1}\left(x-1\right)\)
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 )
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 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
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)
câu a nè = (4x-1)(2x-3)
câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx)
1.
\(3x^2-16x+5\\ =3x^2-x-15x+5\\ =x\left(3x-1\right)-5\left(3x-1\right)\\ =\left(x-5\right)\left(3x-1\right)\)
2.
\(3x^3-14x^2+4x+3\\ =\left(3x^3+x^2\right)-\left(15x^2+5x\right)+\left(9x+3\right)\\ =x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\\ =\left(x^2-5x+3\right)\left(3x+1\right)\)
3. \(x^8+x^7+1\\ =\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\\ =x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x^3-1\right)+x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x\left(x^3+1\right)\left(x+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)[x^2\left(x^3+1\right)\left(x-1\right)+x\left(x^3+1\right)\left(x-1\right)+1]\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+x^5-x^4+x^2-x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)4.
\(64x^4+y^4\\ =\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\\ =\left(8x^2+y^2\right)^2-16x^2y^2\\ =\left(8x^2+y^2-4xy\right)\left(8x^2+y+4xy\right)\)
5.
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\\ =\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\\ =\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+4a^2+2a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)+2a^2\left(x^2+5ax+4a^2\right)+a^4\\ =\left(x^2+5ax+5a^2\right)^2\)
\(A=a^4+a^3+a^3b+a^2b\)
\(A=a\left(a^3+a^2\right)+b\left(a^3+a^2\right)\)
\(A=\left(a+b\right)\left(a^3+a^2\right)\)
\(A=a^2\left(a+1\right)\left(a+b\right)\)
Bài 1:
\(A=x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\)
Đặt \(a=x^2-x+2\) ta có:
\(B=\left(a-1\right).a-12=a^2-a-12=a^2+3a-4a-12=a\left(a+3\right)-4\left(a+3\right)=\left(a+3\right)\left(a-4\right)\)
Thay a = x2 - x + 2 vào ta được:
\(\left(x^2-x+2-4\right)\left(x^2-x+2+3\right)=\left(x^2-x-2\right)\left(x^2-x+5\right)=\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)\)
\(1.4a\left(x-5\right)-2\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(4a+2\right)\)
\(\Leftrightarrow2\left(x-5\right)\left(2a+1\right)\)
\(2.-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(\Leftrightarrow\left(x-3\right)\left(a^2-3a\right)\)
\(\Leftrightarrow a\left(x-3\right)\left(a-3\right)\)
\(3.\text{ }\text{ }\text{ }2a^2b.\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(2a^2b+4a^3b\right)\)
\(\Leftrightarrow2a^2b\left(x+y\right)\left(1+2a\right)\)
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