Question

1 Cho a=\(\frac{-1+\sqrt{2}}{2}\)

b=\(\frac{-1-\sqrt{2}}{2}\)

Tính \(a^7+b^7\)

2 Cho biết \(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}=1\)

Tính \(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\)

Answer 1

2) Dễ thấy\(\left(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=x^2-6x+13-x^2+6x-10=3\)

\(\Leftrightarrow1.\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)=3\)

\(\Leftrightarrow\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}=3\)

Answer 2

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)  = \(\frac{-239}{64}\)