Chứng minh rằng D = x(x-2)(x+3)(5-x)-60<0
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a) \(x^4+2x^3-4x-4=\left[\left(x^2\right)^2-4\right]+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2+2+2x\right)\left(x^2-2\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)=x^2\left(x+1\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) \(x^2+y^2-x^2y^2+xy-x-y=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)=\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(=\left(1-y\right)\left[\left(x-1\right)x+y\left(x-1\right)\left(x+1\right)\right]=\left(1-y\right)\left(x-1\right)\left(x+xy+y\right)\)
c) Không phân tích được.
x4+y4+(x+y)4=x4+y4+x4+4x3y+6x2y2+4xy3+y4
=2x4+2y4+4x2y2+4x3y+4xy3+2x2y2
=2(x4+y4+2x2y2)+4xy(x2+y2)+2x2y2
=2(x2+y2)2+4xy(x2+y2)+2x2y2
=2[(x2+y2)+2xy(x2+y2)+x2y2]
=2(x2+y2+xy)2 (Đpcm)
\(A=\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}=\frac{\left(a+2\right)^2.5a.\left(1-3a\right)}{\left(a-3\right).a.\left(2-a\right)\left(a+2\right)}\)
\(=\frac{\left(a+2\right).5.\left(1-3a\right)}{\left(a-3\right).\left(2-a\right)}\)
ta có:
a^2-1/9=a^2-(1/3)^2=(a+1/3)(a-1/3)
-25+b^2=b^2-5^2=(b+5)(b-5)
K NHA!