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Theo bài ra , ta có :
\(P=\left(\dfrac{x^2}{x^2-y^2}+\dfrac{y}{x-y}\right):\dfrac{x^3-y^3}{x^5-x^4y-xy^4+y^5}\)ĐKXĐ \(x\ne\pm y\)
\(\Leftrightarrow P=\left(\dfrac{x^2}{\left(x-y\right)\left(x+y\right)}+\dfrac{y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}\)
\(\Leftrightarrow P=\left(\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
\(\Leftrightarrow P=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\times\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\Leftrightarrow P=\dfrac{x^4-y^4}{\left(x-y\right)\left(x+y\right)}\)\(\Leftrightarrow P=\dfrac{\left(x^2\right)^2-\left(y^2\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\)(1)
Ta có : \(x+y=5\Rightarrow\left(x+y\right)^2=25\Rightarrow x^2+y^2=25-2xy=25--1=26\)(Vì xy = -1/2)
Thay x2 + y2 = 26 vào (1) ta đk : P = 26
Vậy P = 26 khi x + y = 5 và xy = -1/2
\(P=\left(\dfrac{x^2+y\left(x+y\right)}{\left(x^2-y^2\right)}\right).\left(\dfrac{x^4\left(x-y\right)-y^4\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\\ \)
\(P=\left(\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)}\right).\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x^2+xy+y^2\right)}\)
\(P=x^2+y^2=\left(x+y\right)^2-2xy=25-2\left(-\dfrac{1}{2}\right)=26\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Ta có \(P=\frac{x^2+y\left(x+y\right)}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}\)
\(=\frac{x^2+xy+y^2}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)\(=\frac{x^2+xy+y^2}{x^2-y^2}:\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)}\)
\(=\frac{x^2+xy+y^2}{x^2-y^2}.\frac{\left(x-y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)\(=x^2+y^2=\left(x+y\right)^2-2xy\)
Thay \(x+y=5;xy=-\frac{1}{2}\Rightarrow P=5^2-2.\left(-\frac{1}{2}\right)=26\)
Vậy P=26
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
A=(xy2+xy−x−yx2+xy)(xy2+xy−x−yx2+xy) : (y2x3−xy2+1x+y):xy
A=( \(\dfrac{x}{y\left(x+y\right)}\) - \(\dfrac{x-y}{x\left(x+y\right)}\)) : (\(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}\)+\(\dfrac{1}{x+y}\)) : \(\dfrac{x}{y}\)
A=\(\dfrac{x^2-y\left(x-y\right)}{xy\left(x+y\right)}\) : \(\dfrac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\) : \(\dfrac{y^2-xy+x^2}{x\left(x-y\right)\left(x+y\right)}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\). \(\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{y}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{x}\)
A= 1 - \(\dfrac{y}{x}\)>1
=> y/x <0
=> xy<0 , x+y khác 0
Đặt \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+x+xy+y}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{\left(x+y\right).\left(x-2y\right)}\right):\dfrac{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}{\left(x+y\right).\left(x+1\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left(\dfrac{\left(x-y\right).\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right).\left(2y-x\right)}.\dfrac{\left(x+y\right).\left(x+1\right)}{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}\right):\dfrac{2x^2+y+2}{x+1}\)
\(A=\left(\dfrac{2x^2+y-2}{2y-x}.\dfrac{x+1}{2x^2+y-2}\right).\dfrac{1}{x+1}\)
\(A=\dfrac{1}{2y-x}\)
Thay \(x=-1,76\) và \(y=\dfrac{3}{25}\) vào biểu thức ta được:
\(A=\dfrac{1}{2.\dfrac{3}{25}-\left(-1,76\right)}\)
\(A=\dfrac{1}{2}\)
a/ \(B=\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\left(\dfrac{x+y}{x^2+xy+y^2}+\dfrac{1}{x-y}\right)\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{\left(x+y\right)\left(x-y\right)+x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{x^2-y^2+x^2+xy+y^2}{x^3-y^3}\)
\(=\dfrac{2x^2+xy}{xy}=\dfrac{x\left(2x+y\right)}{xy}=\dfrac{2x+y}{y}\)
b/ Khi x = -1/2 và y = 3 ta có:
\(B=\dfrac{2\cdot\left(-\dfrac{1}{2}\right)+3}{3}=\dfrac{-1+3}{3}=\dfrac{2}{3}\)
\(P=\dfrac{2}{x}-\left(\dfrac{x^2y}{xy\left(x-y\right)}+\dfrac{\left(x^2-y^2\right)\left(x-y\right)}{xy\left(x-y\right)}+\dfrac{xy^2}{xy\left(x-y\right)}\right).\dfrac{x-y}{x^2-xy+y^2}\)
\(P=\dfrac{2}{x}-\left(\dfrac{x^2y+x^3-x^2y-xy^2+y^3+xy^2}{x\left(x-y\right)}\right).\dfrac{x-y}{x^2-xy+y^2}\)\(P=\dfrac{2}{x}-\dfrac{x^3+y^3}{x\left(x-y\right)}.\dfrac{x-y}{x^2-xy+y^2}=\dfrac{2}{x}-\dfrac{\left(x-y\right)\left(x^2-xy+y^2\right)}{x\left(x-y\right)}.\dfrac{x-y}{x^2-xy+y^2}=\dfrac{2}{x}-\dfrac{x-y}{x}=\dfrac{2-x-y}{x}\)Vậy \(P=\dfrac{2-x-y}{x}\)
a. Để x , y xác định thì \(x\ne0\) ; x2 - xy khác 0 ; y2 - xy khác 0 ; x - y khác 0
=> x khác 0; x(x-y) khác 0; xy khác 0 ; y(y-x) khác 0
* Với x(x-y) khác 0 => x khác 0 hoặc x - y khác 0
=> x khác 0 hoặc x khác y
* y(y-x) khác 0 suy ra y khác 0 hoặc y - x khác 0
=> x khác y
Vậy để P xác định thì x và y khác 0 ; và x khác y
P=(\(\dfrac{x^2}{x^2-y^2}+\dfrac{y\left(x+y\right)}{x^2-y^2}\)):\(\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
P=\(\dfrac{X^2+xy+y^2}{x^2-y^2}\).\(\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
P=x^2+y^2=(x+y)^2-2xy=5^2-(-1)=26