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Ta có a.b = 1/4 (1) a - b= -1 (2) a+b = \(\sqrt{2}\) (3) a^6=......=\(\frac{99-70\sqrt{2}}{64}\) (4)
Từ (3) ta có : a^3+b^3 = 2\(\sqrt{2}\)-3a.b(a+b)=\(\frac{5\sqrt{2}}{4}\)
a^6+b^6 = 25.2/16 - 2(1/4)^3 = 99/32
Ta có A=a^7+b^7 = a.a^6 +b.b^6 = a.a^6 +b(99/32-a^6)
=a^6(a-b) +b.99/32= -a^6+b.99/32= \(\frac{99\left(\sqrt{2}+1\right)}{32.2}\)-\(\frac{99-70\sqrt{2}}{64}\)=\(\frac{29\sqrt{2}}{64}\)
a+b=-1; a.b=\(\frac{-1}{4}\) => a2+b2=(a+b)2-2ab=1+\(\frac{1}{2}=\frac{3}{2}\)
a7+b7=(a3+b3)(a4+b4)-a3b3(a+b) (1)
a3+b3=(a+b)((a+b)2-ab))=-1(1+\(\frac{1}{4}\))=\(\frac{-5}{4}\)
a4+b4=(a2+b2)2-2a2b2=\(\left(\frac{3}{2}\right)^2-2.\left(\frac{-1}{4}\right)^2=\frac{17}{8}\)
Thay vào (1) P=\(\frac{-5}{4}.\frac{17}{8}-\left(\frac{-1}{4}\right)^3.\left(-1\right)=\frac{-171}{64}\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
2) Dễ thấy\(\left(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=x^2-6x+13-x^2+6x-10=3\)
\(\Leftrightarrow1.\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=3\)
\(\Leftrightarrow\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}=3\)
Ta có: a+ b= \(\frac{-1+\sqrt{2}}{2}\) + \(\frac{-1-\sqrt{2}}{2}\)= -1
a*b = \(\frac{-1+\sqrt{2}}{2}\)* \(\frac{-1-\sqrt{2}}{2}\)= -\(\frac{1}{4}\)
a2 + b2 = (a+ b)2 - 2ab = 1+ \(\frac{1}{2}\)= \(\frac{3}{2}\)
a4 + b4 = (a2 + b2 )2 - 2a2b2 = \(\frac{9}{4}\)- \(\frac{1}{8}\)= \(\frac{17}{8}\)
a3 + b3 = ( a + b)3 - 3ab(a + b ) = -1-\(\frac{3}{4}\)= \(\frac{-7}{4}\)
vay a7 + b7 = (a3 + b3 )(a4 + b4 ) -a3b3(a+b)= \(\frac{-7}{4}\)* \(\frac{17}{8}\)- (-\(\frac{1}{64}\)) * (-1) = \(\frac{-239}{64}\)
a+b = -1
ab =-1/4
a2+b2=(a+b)2 -2ab = 1+1/2 =3/2
a3+b3 = (a+b)3 - 3ab(a+b) = -1 - 3/4 = -7/4
a4 +b4 = (a2 +b2)2 - 2(ab)2 =9/4 -1/8 = 17/8
a7 +b7 = (a3+b3)(a4 +b4) - a3b3(a+b) = \(-\frac{7}{4}.\frac{17}{8}-\frac{1}{64}.\left(-1\right)=\frac{-238+1}{64}=-\frac{237}{64}\)