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NB
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 3 2019
Áp dụng BĐT: \(\left(a+b\right)^2\ge4ab\) và \(a^2+b^2\ge2ab\) ta có:
\(VT=x^3+4x^2+4x+y^3+4y^2+4y\)
\(VT=x\left(x^2+4x+4\right)+y\left(y^2+4y+4\right)\)
\(VT=x\left(x+2\right)^2+y\left(y+2\right)^2\)
\(\Rightarrow VT\ge x.8x+y.8y=8\left(x^2+y^2\right)\ge16xy\)
\(\Rightarrow VT\ge16xy\)
Dấu "=" xảy ra khi và chi khi \(x=y=2\)
Vậy pt có nghiệm nguyên dương duy nhất \(x=y=2\)
ND
0
AN
0
3 tháng 10 2016
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
L
1
3 tháng 4 2018
\(a,\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-4\left(x^2y^2-4xy-4\right)\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left[2xy+4\right]^2\)
\(=\left(x^2+y^2-5+2xy+4\right)\left(x^2+y^2-5-2xy-4\right)\)
\(=\left[\left(x^2+y^2+2xy\right)-1\right]\left[\left(x^2+y^2-2xy\right)-9\right]\)
\(=\left[\left(x+y\right)^2-1\right]\left[\left(x-y\right)^2-9\right]\)
\(=\left(x+y-1\right)\left(x+y+1\right)\left(x-y-3\right)\left(x-y+3\right)\)
\(b,x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+8x+4\)
\(=x^2\left(x+1\right)+4\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left[\left(x^2+4\right)\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
\(c,x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left[x^2+2x-3x-6\right]\)
\(=\left(x-5\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(d,125x^3-10x^2+2x-1\)
\(=\left(125x^3-1\right)-\left(10x^2-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
23 tháng 11 2017
Ta có :
\(1)\left(x^2+y^2-5\right)-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-[\left(2xy\right)^2+2.2xy.4+4^2]\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left(x^2+y^2-2xy-9\right)\left(x^2+y^2+2xy-1\right)\)
\(=\left[\left(x-y\right)^2-3^2\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y+3\right)\left(x-y-3\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(2)x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-z+z-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-z\right)+x^2y^2\left(z-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y-z\right)\left(x^2y^2-y^2z^2\right)+\left(z-x\right)\left(x^2y^2-z^2x^2\right)\)
\(=\left(y-z\right)\left(xy-yz\right)\left(xy+yz\right)+\left(z-x\right)\left(xy-zx\right)\left(xy+xz\right)\)
\(=y^2\left(y-z\right)\left(x-z\right)\left(x+z\right)+x^2\left(z-x\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(y-z\right)\left(x-z\right)[y^2\left(x+z\right)-x^2\left(y+z\right)]\)
\(=\left(y-z\right)\left(x-z\right)(y^2x+y^2z-x^2y-x^2z)\)
\(=\left(y-z\right)\left(x-z\right)[(y^2x-x^2y)+(y^2z-x^2z)]\)
\(=\left(y-z\right)\left(x-z\right)[xy(y-x)+z(y^2-x^2)]\)
\(=\left(y-z\right)\left(x-z\right)[xy(y-x)+z(y-x)\left(x+y\right)]\)
\(=\left(y-z\right)\left(x-z\right)(y-x)\left(xy+xz+yz\right)\)
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