x^2-2x-4xy^2-4y
Rút gọn: \(\frac{2x^2-4xy}{x^2+4xy+4y^2}:\frac{4y^2-x^2}{x^2-4xy+4y^2}:\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)
Rút gọn: \(\frac{2x^2-4xy}{x^2+4xy+4y^2}:\frac{4y^2-x^2}{x^2-4xy+4y^2}:\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^3}\)
\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)
\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)
\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)
min x^2 + 4y^2 - 4xy + 2x - 4y + 9
Đặt \(P=x^2+4y^2-4xy+2x-4y+9\)
\(P=\left(x-2y\right)^2+2\left(x-2y\right)+1+8\)
\(P=\left(x-2y+1\right)^2+8\ge8\)
\(P_{min}=8\) khi \(x-2y+1=0\)
y^2-9-x^2+6x
25-4x^2-4xy-y^2
x^2-xz+4y^2-2yz+4xy
3x^2+6xy-48z^2+3y^2
x^2-z^2+4y^2-4t^2-4xy+4zt
x^3+2x^2y+xy^2-16x
Rút gọn:
\(\frac{2x^2-4xy}{x^2+4xy+4y^2}\): \(\frac{4y^2-x^2}{x^2-4xy+4y^2}\): \(\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^2}\)
D=x^2+4y^2-2x+10+4xy-4y tại x+2y=5
\(D=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4y^2+4xy\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay \(x+2y=5;\)có :
\(D=5^2-2.5+10\)
\(=25-10+10\)
\(=25\)
Vậy...
cho 2x+y=5 tính D = x^2+4xy-2x+10+4y^2-4y
1,Tìm GTNN
\(2x^2+5y^2-4xy-2x+4y+10\)
2,Tìm GTLN
a,\(3-10x^2-4xy-4y^2\)
b,\(-x^2-y^2+2x-4y-4\)
1) (x-1)2 + (x- 4y)2 + (y + 2)2 +10 -1-4
GTNN = 5
2) tuong tu
CHUNG MINH x^2 + 4y^2 -2x - 4xy + 4y + 2018 > 0