a) Ta có:
\(x+y=-1\)
\(\Rightarrow\left(x+y\right)^2=\left(-1\right)^2\)
\(\Rightarrow x^2+y^2+2xy=1\)
Thay xy = -6 vào ta được
\(x^2+y^2+2.\left(-6\right)=1\)
\(\Rightarrow x^2+y^2-12=1\)
\(\Rightarrow x^2+y^2=1+12\)
\(\Rightarrow x^2+y^2=13\)
b) Ta có:
\(x+y=17\)
\(\Rightarrow\left(x+y\right)^2=17^2\)
\(\Rightarrow x^2+y^2+2xy=289\)
Thay xy = 72 vào ta được:
\(x^2+y^2+2.72=289\)
\(\Rightarrow x^2+y^2+144=289\)
\(\Rightarrow x^2+y^2=289-144=145\)
Ta lại có:
\(\left(x-y\right)^2\)
\(=x^2+y^2-2xy\)
Thay x2 + y2 = 145 và xy = 72
\(=145-2.72\)
\(=145-144\)
\(=1\)
c) Ta có:
\(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Rightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)