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Cách làm tổng quát của những bài như thế này là em đưa về hằng đẳng thức:
Đối với những bài có hệ số ở x^2 em đưa ra ngoài:
\(-3x^2+5x+4\)
\(=-3\left(x^2-\frac{5}{3}x-\frac{4}{3}\right)\)
\(=-3\left(x^2-2.x.\frac{5}{2.3}+\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^2-\frac{4}{3}\right)\)
\(=-3\left(x^2-2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2-\frac{73}{36}\right)\)
\(=-3\left(x-\frac{5}{6}\right)^2+3.\frac{73}{36}\)
\(=3\left(\frac{73}{36}-\left(x-\frac{5}{6}\right)^2\right)=3\left(\frac{\sqrt{73}}{6}-x+\frac{5}{6}\right)\left(\frac{\sqrt{73}}{6}+x-\frac{5}{6}\right)\)
\(=3\left(\frac{\sqrt{73}+5}{6}-x\right)\left(\frac{\sqrt{73}-5}{6}+x\right)\)
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\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
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1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
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a) 49( y - 4 )2 - 9( y + 2 )2
= 72( y - 4 )2 - 32( y + 2 )2
= ( 7y - 28 )2 - ( 3y + 6 )2
= ( 7y - 28 - 3y - 6 )( 7y - 28 + 3y + 6 )
= ( 4y - 34 )( 10y - 22 )
= 2( 2y - 17 ).2( 5y - 11 )
= 4( 2y - 17 )( 5y - 11 )
b) 3x2 - 5x - 2 = 3x2 - 6x + x - 2 = 3x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 3x + 1 )
c) x2 - 7xy + 10y2 = x2 - 2xy - 5xy + 10y2 = x( x - 2y ) - 5y( x - 2y ) = ( x - 2y )( x - 5y )
d) 3x2 - 10xy + 3y2 = 3x2 - 9xy - xy + 3y2 = 3x( x - 3y ) - y( x - 3y ) = ( x - 3y )( 3x - y )
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\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+1+2\left(x+y\right)\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2+2xy+1+2x+2y-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy+1+2x+2y\right)\)
\(=\left(x+y-1\right)\left[\left(x^2+1+2x\right)\left(y^2-xy+2y\right)\right]\)
\(=\left(x+y-1\right)\left(x+1\right)^2\left(y-x+2\right)y\)
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ai giup minh vs
C=(1/2+1/3+....+1/2017+1/2018)/(1/2017+2/2016+.....+2016/2+2017/1)
co gang giup minh :( minhthanks nhieu lam
\(16x-5x^2-3=\left(-5x^2+15x\right)+\left(x-3\right)=-5x.\left(x-3\right)+\left(x-3\right)\\ \)
\(=\left(x-3\right).\left(1-5x\right)\)
\(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x+2\right)\)
\(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-\left(5x^2-16x+3\right)\)
\(=-\left(5x^2-15x-x+3\right)\)
\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)
\(=-\left[\left(5x-1\right)\left(x-3\right)\right]\)
Câu 2:
\(x^2-x-6\)
\(=x^2+2x-3x-6\)
\(=x\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x-3\right)\left(x+2\right)\)