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a, \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)
\(=\left(2a-3b\right)\left[\left(2a\right)^2+2a.3b+\left(3b\right)^2\right]-2a\left(2a-3b\right)\left(2a+3b\right)\)
\(=\left(2a-3b\right)\left[4a^2+6ab+9b^2-2a\left(2a+3b\right)\right]\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)
\(=\left(2a-3b\right).9b^2\)
b, \(\left(x^3-y^3\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+x-y\right)\)
c, \(\left(m^3+n^3\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2+m+n\right)\)
Chúc bạn học tốt!!!
a) 5ay - 3bx + ax - 15by
= (5ay + ax) - (3bx + 15by)
= a (5y + x) - 3b (x + 5y)
= (5y + x) (a - 3b)
b) x^3 + x^2 - x - 1
= (x^3 + x^2) - (x + 1)
= x^2 (x + 1) - (x + 1)
= (x + 1) (x^2 - 1)
c) (2a + b)^2 - (2b + a)^2
= 4a^2 + 4ab + b^2 - 4b^2 - 4ab - a^2
= 3a^2 - 3b^2
= 3 (a^2 - b^2)
d) (8a^3 - 27b^3) - 2a (4a^2 - 9b^2)
= 8a^3 - 27b^3 - 8a^3 + 18ab^2
= 27b^3 + 18ab^2
= 9b^2 (3b + 2a)
a) 4a2b3 - 6a3b2 = 2a2b2( 2b - 3a )
b) ( a - b )2 - ( b - a ) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
c) ( 8a3 - 27b3 ) - 2a( 4a2 - 9b2 ) = 8a3 - 27b3 - 8a3 + 18ab2 = 18ab2 - 27b3 = 9b2( 2a - 3b )
d) 10x2 + 10xy + 5x + 5y = 10x( x + y ) + 5( x + y ) = ( x + y )( 10x + 5 ) = 5( x + y )( 2x + 1 )
e) 5ay - 3bx + ax - 15by = 5y( a - 3b ) + x( a - 3b ) = ( a - 3b )( 5y + x )
a) \(4a^2.b^3-6a^3.b^2=2a^2.b^2\left(2b-3a\right)\)
b) \(\left(a-b\right)^2-\left(b-a\right)=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right).\left(a-b+1\right)\)
c) \(8a^3-27b^3-2a.\left(4a^2-9b^2\right)=8a^3-27b^3-8a^3+18ab^2\)
\(=-27b^3+18ab^2=18ab^2-27b^3=9b^2.\left(2a-3b\right)\)
d) \(10x^2+10xy+5x+5y=5.\left(2x^2+2xy+x+y\right)\)
\(=5.\left[\left(2x^2+2xy\right)+\left(x+y\right)\right]=5.\left[2x\left(x+y\right)+\left(x+y\right)\right]\)
\(=5\left(x+y\right)\left(2y+1\right)\)
e) \(5ay-3bx+ax-15by=\left(5ay-15by\right)-\left(3bx-ax\right)\)
\(=5y\left(a-3b\right)-x\left(3b-a\right)=5y\left(a-3b\right)+x\left(a-3b\right)\)
\(=\left(a-3b\right)\left(x+5y\right)\)
a) \(N=8a^3-27b^3\)
\(=\left(2a\right)^3-\left(3b\right)^3\)
\(=\left(2a-3b\right)^3+18ab\left(2a-3b\right)\)
\(=5^3+18\cdot12\cdot5\)
\(=125+1080=1205\)
b) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+2ab+b^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a+b\right)^2\)
\(=\left(a+b\right)^3+3ab\left(a+b\right)\left(a+b-1\right)\)
\(=1^3+3ab\cdot1\cdot0\)
\(=1\)
a ) \(N=8a^3-27b^3\)
\(\Leftrightarrow N=\left(2a-3b\right)\left(4x^2+6ab+9b^2\right)\)
\(\Leftrightarrow N=5\left(4x^2+9b^2+72\right)\)
Ta có : \(2a-3b=5\)
\(\Leftrightarrow4a^2+9b^2=25+6ab\)
Thay vào ta được : \(N=5\left(25+6ab+72\right)=845\)
b ) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(\Leftrightarrow K=\left(a+b\right)^3-3ab\left(a+b\right)+6a^2b^2\left(a+b\right)+3ab\left(a+b\right)^2-6a^2b^2\)
\(\Leftrightarrow K=1-3ab+6a^2b^2+3ab-6a^2b^2=1\)
c ) \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)
\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)
\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)
\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)
\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)
a) Ta có: 27\(x^3\)+ y\(^3\) = (3x)\(^3\) + y\(^3\)= (3x + y)[(3x)\(^2\) – 3x . y + y\(^2\)] = (3x + y)(9x\(^2\) – 3xy + y\(^2\))
Nên: (3x + y) (9x\(^2\) – 3xy + y\(^2\)) = 27x\(^3\) + y\(^3\)
b) Ta có: 8x\(^3\) – 125 = (2x)\(^3\) – 53= (2x – 5)[(2x)\(^2\) + 2x . 5 + 5\(^2\)]
= (2x – 5)(4x\(^2\) + 10x + 25)
Nên:(2x – 5)(4x\(^2\) + 10x + 25)= 8x\(^3\) – 125
Trả lời:
a) Ta có:
27x3 + y3 = (3x)3 + y3= (3x + y)[(3x)2 – 3x . y + y2] = (3x + y)(9x2 – 3xy + y2)
Nên: (3x + y) (9x2 – 3xy + y2 ) = 27x3 + y3
b) Ta có:8x3 - 125 = (2x)3 - 53= (2x - 5)[(2x)2 + 2x . 5 + 52]
= (2x - 5)(4x2 + 10x + 25)
Nên: (2x - 5)(4x2+ 10x +25 ) = 8x3 - 125
a: \(=a^2-b^4\)
b: \(=\left(a^2+2a\right)^2-9\)
c: \(=a^2-\left(2a+3\right)^2\)
d: \(=a^4-\left(2a-3\right)^2\)
e: \(=\left(-a^2-2a+3\right)^2\)
g: \(=4a^2-a^4\)
Lời giải:
Dựa vào công thức hằng đẳng thức đáng nhớ:
\(x^3+y^3=(x+y)(x^2-xy+y^2)\)
\(x^3-y^3=(x-y)(x^2+xy+y^2)\)
Ta có thể điền như sau:
\((2a+3b)(4a^2-6ab+9b^2)=8a^3+27b^3\)
\((5x-4y)(25x^2+20xy+16y^2)=(5x)^3-(4y)^3=125x^3-64y^3\)