Ta có:\(\left(x-y\right)^2+2xy=x^2-2xy+y^2+2xy=x^2+y^2\)

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

\(=7^2+2.60=49+120=169\)

\(A=\left(x-y\right)\left(x+y\right)=7\left(x+y\right)\)

Có \(\left(x-y\right)^2=49\)

\(\Leftrightarrow x^2+y^2-2xy=49\)

\(\Leftrightarrow\left(x^2+y^2+2xy\right)-4xy=49\)

\(\Leftrightarrow\left(x+y\right)^2=289\)

\(\Leftrightarrow x+y=17\)

\(\Rightarrow A=7.17=119\)

Vậy ....

Ta có: x - y = 7

<=> (x-y)²=49

<=>x²-2xy+y²=49

<=>x²+y²-2.60=49

<=>x²+y²-120=49

<=>x²+y²=49+120

<=>x²+y²=169

+)Ta có: x²+y²=169 (câu a) 

<=> (x+y)²-2xy=169

<=>(x+y)²=169+2xy=169+2.60=289

<=>x+y=17

=>\(C=x^2-y^2=\left(x-y\right)\left(x+y\right)=7.17=119\)

+) x²+y²=169 

<=>(x²+y²)²=169²

<=>x⁴+2x²y²+y⁴=28561

<=>x⁴+y⁴=28561-2(xy)²=28561-2.60²=28561-7200=21361

x - y = 7 => y = x - 7

=> x(x - 7) = 60

     x² - 7x + 12,25 = 72,25

   (x - 3,5)²  = 72,25 mà x > 0 => x - 3,5 > -3,5

=> x - 3,5 = 8,5 => x = 12 => y = 60 : 12 = 5 => P = 12⁴ - 5⁴ = 20111

cảm ơn bạn! bạn có thể trả lời câu hỏi nữa mk vừa đăng lên ko

ta có:x.y=60 =>(xy)^2=60^=3600

(x-y)^2=x²-2xy+y^2=7^2

          =>x^2+y^2=49+2*60=169  

(x²-y²)²=x⁴-2*x²y²+y⁴=169²

            =>x⁴-y⁴=169²+2*3600=21361

\(C=x^2-y^2\)

Tương tự câu \(A=x^2+y^2\)

\(D=x^4+y^4\)

Thay x + y = 17; x.y = 60 vào \(\left(x+y\right)^2=x^2+2xy+y^2\):

17² = x² + 2.60 + y²

289 = x² + 120 + y²

\(\Leftrightarrow x^2+y^2=169\)

Lại có:

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

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

Thay \(x^2+y^2=169;x.y=60\)vào biểu thức trên:

169² = x⁴ + y⁴ + 2 . 60²

\(\Leftrightarrow x^4+y^4=28561-7200\)

\(\Leftrightarrow x^4+y^4=21361\)

C1: Ta có: \(x-y=7\Leftrightarrow\left(x-y\right)^2=49\Leftrightarrow x^2-2xy+y^2=49\Leftrightarrow x^2+y^2=49+2xy=49+2.60=169\)

=>\(B=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=7\left(169+60\right)=7.229=1603\)

C2: \(B=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left[\left(x-y\right)^2+3xy\right]=7\left(7^2+3.60\right)=7.229=1603\)

Ta có: \(x-y=7\)

\(\Leftrightarrow\left(x-y\right)^2=7^2\)

\(\Leftrightarrow x^2-2xy+y^2=49\)

\(\Leftrightarrow x^2+2xy+y^2-4xy=49\)

\(\Leftrightarrow\left(x+y\right)^2-4\cdot60=49\) (vì \(xy=60\))

\(\Leftrightarrow\left(x+y\right)^2=49+240\)

\(\Leftrightarrow\left(x+y\right)^2=289\)

\(\Rightarrow x+y=17\) (vì \(x>y>0\))

Mặt khác: \(x^2-y^2\)

\(=\left(x-y\right)\left(x+y\right)\)

\(=7\cdot17\) (vì \(x-y=7;x+y=17\))

\(=119\)

#Urushi☕

Ta có: 

\(x-y=7\)

\(\Leftrightarrow y=x-7\) (1)

Mà: \(xy=60\) (2)

Thay (1) vào (2) ta có: 

\(x\cdot\left(x-7\right)=60\) (ĐK: \(x>y>0\))

\(\Leftrightarrow x^2-7x=60\)

\(\Leftrightarrow x^2-7x-60=0\)

\(\Leftrightarrow x^2+5x-12x-60=0\)

\(\Leftrightarrow x\left(x+5\right)-12\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-12\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\)

Ta có: \(x=12\)

\(\Leftrightarrow y=12-7=5\)

Giá trị của bt là:

\(12^2-5^2=144-25=119\)

\(P=\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}+2020=\dfrac{x^5+y^5}{\left(xy\right)^2}+2020=\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)-\left(xy\right)^2\left(x+y\right)}{\left(-2\right)^2}\)

\(=\dfrac{\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\left[\left(x+y\right)^2-2xy\right]-\left(-2\right)^2.5}{4}\)

\(=\dfrac{\left(-8+6.5\right)\left(25+4\right)-20}{4}=...\)

\(B=8x^2+2x-8x^3-8x^2+8x^3-2x+3=3\)

\(C=x^3-3x^2+3x-1+x^3+3x^2+3x+1+2x^3-8x=4x^3-2x\)

\(D=\left(x+y-5\right)^2-2\left(x+y-5\right)\left(x+3\right)+\left(x+3\right)^2=\left(x+y-5-x-3\right)^2=\left(y-8\right)^2\)

câu 2. ta có 

a.\(\left(x-y\right)^2=\left(x+y\right)^2-4xy=7^2-4\times12=1\)

b.\(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)=3\left(x+y\right)^2-6xy-2\left(x+y\right)^3+6xy\left(x+y\right)=3-6xy-2+6xy=1\)