\(\frac{5x^2-10xy+5^2}{2^2-2xy+2y^2}:\frac{8y-8x}{x^2-xy+y^2}\)
Những câu hỏi liên quan
\(\frac{5x^2-10xy+5y^2}{2x^2-2xy+2^2}:\frac{8y-8x}{x^2-xy+y^2}\)
Rút gọn biểu thức :
a) \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}\)
b) \(\dfrac{5x^2-10xy+5y^2}{2x^2-2xy+2y^2}:\dfrac{8x-8y}{10x^3+10^3}\)
Tính :
a)\(\dfrac{6x-3}{5x^2+x}.\dfrac{25x^2+10x+1}{1-8x^3}\)
b)\(\dfrac{3x^2-x}{x^2-1}.\dfrac{1-x^4}{\left(1-3x\right)^3}\)
c)\(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}\)
d) \(\dfrac{5x^2-10xy+5y^2}{2x^2-2xy+2y^2}:\dfrac{8x-8y}{x^3+10y^3}\)
P = x^2 +2y^2 - 2xy + 8x + 8y + 2017
Q = - x^2 - 2y^2 - 2xy + 8x + 6y + 13
E = -x^2 - 4y^2 + 2xy + 2x + 10xy - 3
tìm giá trị lớn nhất
P=\(X^2+2Y^2-2XY+8X+8Y+2017\)
P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)
P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)
DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)
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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: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)
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Rút gọn: \(\frac{2x^2-4xy}{x^2+4xy+4y^2}:\frac{4y^2-x^2}{x^2-4xy+4y^2}:\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^3}\)
\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)
\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)
\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)
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1. Rút gọn biểu thức x(x-y)-y(x+y)+x^2+y^2
2. Phân tích đa thức thành nhân tử :
a) a^3-a^2x-ay^2+xy^2
b) 5x^2-4x+10xy
c) 12x-9--4x^2
d) 8x^3+12x^2y+6xy^2+y^3
e) 5x^2-4x+10xy-8y
3. Điền vào chỗ trống :
a) (1/2x-y)^2=1/4x^2-.....+y^2
Rút gọn:
\(\frac{2x^2-4xy}{x^2+4xy+4y^2}\): \(\frac{4y^2-x^2}{x^2-4xy+4y^2}\): \(\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^2}\)
Bài 1: Phân tích đa thức thành nhân tử
1. 5x-10-xy+2y
2.2x^2+2y^2-4xy-xz+yz
3.5x^2y-10xy^2
4.3x^2-6xy+3y^2-12z^2
5.x^2+4xy-16+4y^2
6.7x-6x^2-2
7.(2x+y)^2+x(2x+y)
8.x(x-y)+5x-5y
9.x^2-y^2+2x+1
10.x^3-9x
11.xy-2y+x-2
12.x^3-3x^2-4x+12
13.3x-x^2-2xy+3y-y^2
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
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