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

P = ( xy + 1 ) ( x²y² - xyt + 1 )

   = x³y³ + 1

   = \(\left(5.\frac{3}{5}\right)^3+1\)

   = \(27+1\)

    = 28

=28

tính r

a) \(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2.4=25-8=17\)

b) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.4.5=125-60=65\)

c) \(x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2\)

\(=\left(\left(x+y\right)^2-2xy\right)^2-2\left(xy\right)^2=\left(5^2-2.4\right)^2-2.4^2\)

\(=\left(25-8\right)^2-2.16=17^2-32=289-32=257\)

d) \(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)

\(=\left(x+y\right)^5-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(=\left(x+y\right)^5-5xy\left(\left(x^3+y^3\right)+\left(2x^2y+2xy^2\right)\right)\)

\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(2xy\left(x+y\right)\right)\right)\)

\(=\left(5\right)^5-5.4\left(\left(\left(5^3-3.4.5\right)+\left(2.4.5\right)\right)\right)\)

\(=3125-20\left(125-65+40\right)\)

\(=3125-20\left(100\right)=3125-2000=1125\)

\(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2\cdot4=25-8=17\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot4\cdot5=125-60=65\\ x^4+y^4 \\ =\left(x+y\right)^4-4xy\left(x^2+y^2\right)-6x^2y^2\\ =5^4-4\cdot4\left[\left(x+y\right)^2-2xy\right]-6\left(xy\right)^2\\ =5^4-4\cdot4\cdot\left(5^2-2\cdot4\right)-6\cdot4^2\\ =625-16\cdot\left(25-8\right)-6\cdot16\\ =625-16\cdot17-96\\ =625-272-96\\ =257\\ x^5+y^5\\ =\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\\ =5^5-5\cdot4\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-10\left(xy\right)^2\cdot5\\ =3125-20\left(5^3-3\cdot4\cdot5\right)-10\cdot4^2\cdot5\\ =3125-20\cdot\left(125-60\right)-10\cdot16\cdot5\\ =3125-20\cdot65-800\\ =3125-1300-800\\ =1025\)

rút gọn P đi

26 mk vừa làm xong

Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\)(bđt cosi)

=> \(\frac{\left(x+y\right)^2}{4}\ge4\) <=> \(\left(x+y\right)^2\ge16\) <=> \(x+y\ge4\)

CM bđt tương đương: \(\frac{1}{x+3}+\frac{1}{y+3}\le\frac{2}{5}\) 

<=> \(\frac{5\left(x+3\right)+5\left(y+3\right)}{\left(y+3\right)\left(y+3\right)}\le2\)

<=> \(2\left(xy+3x+3y+9\right)\ge5x+5y+30\)

<=> \(2.4+6\left(x+y\right)+18-5\left(x+y\right)-30\ge0\)

<=> \(x+y-4\ge0\) (vì x + y \(\ge\)4)

<=> \(4-4\ge0\) (Luôn đúng) 

=> ĐPCM

* Với M

Ta có M= x²+y² = x²+y²+2xy-2xy=(x+y)² - 2xy= (-9)² - 2.18 = 81- 36 = 45

* Với N 

Ta có M = x⁴ + y⁴ = (x²)² + (y²)² + 2(xy)² - 2(xy)² = (x²+y²)² + 2 (xy)²= 45² + 2. 18²= 2673

* Với T 

Ta có T = x² - y²  => chịu

x^2 +y^2 =x^2 + 2xy + y^2 - 2xy

(x+y)^2 - 2xy

(-9)^2-2*18

81 - 36

45