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\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)
\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)
1) -25x6 - y8 + 10x3y4 = -( 25x6 - 10x3y4 + y8 ) = -[ ( 5x3 )2 - 2.5x3.y4 + ( y4 ) ] = -( 5x3 - y4 )2
2) 2x( 3x - 5 ) + 10 - 6x = 2x( 3x - 5 ) - 2( 3x - 5 ) = ( 3x - 5 )( 2x - 2 ) = 2( 3x - 5 )( x - 1 )
3) x2 - 9 - x2( x2 - 9 ) = ( x2 - 9 ) - x2( x2 - 9 ) = ( x2 - 9 )( 1 - x2 ) = ( x - 3 )( x + 3 )( 1 - x )( 1 + x )
4) 4x2 - 9 - ( 3x + 1 )( 2x - 3 ) = ( 2x - 3 )( 2x + 3 ) - ( 3x + 1 )( 2x - 3 )
= ( 2x - 3 )[ ( 2x + 3 ) - ( 3x + 1 ) ]
= ( 2x - 3 )( 2x + 3 - 3x - 1 )
= ( 2x - 3 )( 2 - x )
5) 8x3 - y3 - 4x + 2y = ( 8x3 - y3 ) - ( 4x - 2y )
= [ ( 2x )3 - y3 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 - 2 )
1) \(=\left(2z+3\right)\left(4z^2-6z+9\right)\)
2) \(=\left(\frac{3x^2}{5}-\frac{1}{2}\right)\left(\frac{3x^2}{5}+\frac{1}{2}\right)\)
3) \(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
4) \(=\left(2x+1\right)^2\)
5) \(=\left(x-10\right)^2\)
6) \(=\left(y^2-7\right)^2\)
7) \(=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
a) 1/2(x3+8)=1/2(x+2)(x2-2x+4)
b) x4(x-y)+2x3(x-y)=x3(x+2)(x-y)
c) x2-(y2-6y+9)=x2-(y-3)2=(x-y+3)(x+y-3)
d) xy(x3+y3)=xy(x+y)(x2-xy+y2)
e)3x2(x2-25y2)=3x2(x-5y)(x+5y)
f) 4x4+4x2y2+y4-4x2y2= (2x2+y2)2-(2xy)2=(2x2-2xy+y2)(2x2+2xy+y2)
a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)
b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)
c) \(x^2-y^2+6y-9=x^2-\left(y-3\right)^2=\left(x+y-3\right)\left(x-y+3\right)\)
d) \(x^4y+xy^4=xy\left(x^3+y^3\right)=xy\left(x+y\right)\left(x^2-xy+y^2\right)\)
e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).
f) \(4x^4+y^4=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2+2xy\right)\left(2x^2-y^2-2xy\right)\)
mình chỉ phân tích thôi
a) 6x(4-x)+x-4
=6x(4-x)-(4-x)
=(6x-1)(4-x)
c) 25x^2-10x+1-16z^2
=(5x-1)^2-16z^2
=(5x-1-4z)(5x-1+4z)
ban xem lại đề bài câu b đi chắc là sai đó
còn các câu trên bạn tự làm nhé
Thực hiện phép tính:
a) (2x-3y)(4x2+6xy+9y2)
=8x3-27y3
b) (6x3+3x2+4x+2):(3x2+2)
=(3x2+2)(2x+1):(3x2+2)
=2x+1
c) (x+2)2+(3-x)-2(x+3)(x-3)
=x2+4x+4+3-x-2x2+18
=-x2+4x+25
\(x^2-6x+9=\left(x-3\right)^2\)
\(25+10x+x^2=\left(5+x\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a+2b^2\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}-y^4\right)^2\)
\(\left(3x+2\right)^2-4=\left(3x+2-2\right)\left(3x+2+2\right)=3x\left(3x+4\right)\)
\(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
\(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
\(\frac{9}{25}x^4-\frac{1}{4}=\left(\frac{3}{5}x^2-\frac{1}{2}\right)\left(\frac{3}{5}x^2+\frac{1}{2}\right)\)