Ta có: \(A\cdot1=\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\right)=x^2-6x+13-x^2+6x-10=3\)

=> A = 3

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

\(\Rightarrow\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right).1=3\)

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)  = \(\frac{-239}{64}\)

Ta có :

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

=\(\sqrt{x^2-2.3.x+3^2+4}-\sqrt{x^2-2.3.x+3^2+1}\)

=\(\sqrt{\left(x-3\right)^2+2^2}-\sqrt{\left(x-3\right)^2+1^2}\)

Ta có :

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

\(=\sqrt{x^2-6x+9+4}+\sqrt{x^2-6x+9+1}\)

\(=\sqrt{\left(x-3\right)^2+2^2}+\sqrt{\left(x-3\right)^2+1}\)

đề bài thiếu rồi bạn ơi

\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)

\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)

\(\Rightarrow T=3\)

Đặt \(a=\sqrt{x^2-6x+19},a\ge0\) ; \(b=\sqrt{x^2-6x+10},b\ge0\)

\(\Rightarrow\begin{cases}a-b=3\\a^2-b^2=9\end{cases}\)  \(\Rightarrow A=a+b=3\)

bạn kiểm tra lại biểu thức A đi bạn