Câu 1: 

a: \(\Leftrightarrow2x^2-x-5< x^2+x-6\)

\(\Leftrightarrow x^2-2x+1< 0\)

hay \(x\in\varnothing\)

b: \(\Leftrightarrow x^2-5x-x+4>0\)

\(\Leftrightarrow x^2-6x+4>0\)

\(\Leftrightarrow\left(x-3\right)^2>5\)

hay \(\left[{}\begin{matrix}x>\sqrt{5}+3\\x< -\sqrt{5}+3\end{matrix}\right.\)

( x+  y)^2 = 3^2

=> x^2 + 2xy + y^2 = 9 

=> x^2 + y^2 = 9 - 2xy = 9-2.2 = 9 - 4 = 5 

Vậy A = 5 

b) ( x^2 + y^2 )^2 = 5^2 

=> x^4 + y^4 + 2x^2y^2 = 25

=> x^4 + y^4 = 25 - 2x^2y^2 

=> x^4 + y^4 = 25 - 2(xy)^2 

                   = 25 - 2 (2)^2 = 25 - 2.4 = 25 - 8 = 17 

ta có x-y = 7 => (x-y)^2 = 49 <=> x^2 + y^2 - 2xy = 49 <=> x^2+y^2 - 2*60 = 49 <=> x^2+y^2 = 49+ 120 <=> x^2+y^2 = 169 => \(\left(x^2+y^2\right)^2=169^2\)<=> x^4+y^4 + \(2x^2y^2\)= 28561 (1)

từ xy = 60 => x^2 * y^2 = 360 => 2x^2 * y^2 = 720 thay vào 1 tính được A= x^4 + y^4 = 27841 

Chúc bạn học tốt!

to cam on

a/ \(M=x^4-xy^3+x^3y-y^4-1\)

\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)

Mà \(x+y=0\)

\(\Leftrightarrow M=x^3.0-y^3.0-1\)

\(\Leftrightarrow M=-1\)

Vậy ...

A=(1/2+-1)*(1/2^2-1/2*-1+-1^2)

A=-1/2*-1/4

A=1/8

a: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{5}{-2}=-\dfrac{5}{2}\)

b: \(x^2+y^2=\left(x+y\right)^2-2xy=25-2\cdot\left(-2\right)=29\)

c: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{x^2+y^2}{\left(xy\right)^2}=\dfrac{29}{4}\)

d: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot\left(-2\right)\cdot5=125+6\cdot5=155\)