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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)\(N=\left(\frac{x^2}{x^2-y^2}+\frac{y}{x-y}\right):\frac{x^3-y^3}{x^5-x^4y-xy^4+y^5}\)
\(=\left(\frac{x^2}{\left(x-y\right)\left(x+y\right)}+\frac{xy+y^2}{\left(x-y\right)\left(x+y\right)}\right):\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x^4-y^4\right)\left(x-y\right)}\)
\(=\frac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}:\frac{\left(x^2+xy+y^2\right)}{x^4-y^4}\)
\(=\frac{x^4-y^4}{\left(x-y\right)\left(x+y\right)}\)
\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{x^2-y^2}=x^2+y^2\)
b) Ta có: \(x+y=\frac{1}{40}\)
\(\Rightarrow\left(x+y\right)^2=\frac{1}{1600}\)
\(\Rightarrow x^2+2xy+y^2=\frac{1}{1600}\)
\(\Rightarrow x^2-\frac{1}{40}+y^2=\frac{1}{1600}\)
\(\Rightarrow x^2+y^2=\frac{1}{1600}+\frac{1}{40}\)
\(\Rightarrow x^2+y^2=\frac{41}{1600}\)
Vậy \(N=\frac{41}{1600}\)
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
\(P=\left(\dfrac{x^2+y\left(x+y\right)}{\left(x^2-y^2\right)}\right).\left(\dfrac{x^4\left(x-y\right)-y^4\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\\ \)
\(P=\left(\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)}\right).\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x^2+xy+y^2\right)}\)
\(P=x^2+y^2=\left(x+y\right)^2-2xy=25-2\left(-\dfrac{1}{2}\right)=26\)
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
rút gọn P đi
26 mk vừa làm xong