Ta có x=7+y thay vào x.y=60 ta được (7+y).y=60 =>y=-12 , x=-5

a)x²-y²=(-12)²-(-5)²=119

b)x⁴+y⁴=(-12)⁴+(-5)⁴=21361

có hệ thức viet nhanh hơn mà mình quên rồi :)) nhớ nhe

Ta có : x + y = 7 

=> y = 7 - x 

Lại có : x(7 - x) = 12

<=> 7x - x² = 12

<=> 7x - x² - 12 = 0

<=> x² - 7x + 12 = 0 (chia cả hai vế cho -1)

<=> x² - 3x - 4x + 12 = 0 

<=> x(x - 3) - 4(x - 3) = 0 

<=> (x - 3)(x - 4) = 0 

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=4\left(tm\right)\end{cases}}}\)

=> y = 4 ; 3 

Ta có : |x - y| = |3 - 4| = 1 

           |x - y| = |4 - 3| = 1

TA có x+y=7 suy ra x=7-y

Thay vào bt x.y=12 ta có \(\left(7-y\right)y=12\Rightarrow7y-y^2=12\)

\(\Rightarrow y^2-7y+12=0\)

\(\Rightarrow\left(y-3\right)\left(y-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y-3=0\\y-4=0\end{cases}\Rightarrow\orbr{\begin{cases}y=3\Rightarrow x=4\Rightarrow\left|x-y\right|=1\\y=4\Rightarrow x=3\Rightarrow\left|x-y\right|=1\end{cases}}}\)

Vậy \(\left|x-y\right|=1\)

\(A=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.5.4=65\)

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

a) − 3 ( x + 7 ) 2 x + 1                b)  y ( y 2 + 1 ) ( 3 y − 2 ) 2

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

b)x⁴+y⁴

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

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

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

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