Bài 1:
\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{8-2\sqrt{8.9}+9}=\sqrt{(\sqrt{8}-\sqrt{9})^2}\)
\(=|\sqrt{8}-\sqrt{9}|=3-2\sqrt{2}\)
\(\Rightarrow a=3; b=-\sqrt{2}\)
\(\Rightarrow a^2+b^2=9+2=11\)
Bài 1:
Ta có: \(\sqrt{17-12\sqrt{2}}=a+b\sqrt{2}\)
\(\Leftrightarrow a+b\sqrt{2}=3-2\sqrt{2}\)
Suy ra: a=3; b=-2
\(\Leftrightarrow a^2+b^2=3^2+\left(-2\right)^2=9+4=13\)
Bài 2:
a) Ta có: \(\dfrac{\sqrt{a}-1}{a\sqrt{a}+\sqrt{a}-a}:\dfrac{1}{a^2+a}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{a\left(a+1\right)}{1}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+1\right)}{\left(a-\sqrt{a}+1\right)}\)
Bài 2:
ĐKXĐ: $a>0$
\(\frac{\sqrt{a}-1}{a\sqrt{a}+\sqrt{a}-a}:\frac{1}{a^2+a}=\frac{\sqrt{a}-1}{\sqrt{a}(a-\sqrt{a}+1)}.a(a+1)=\frac{(a-\sqrt{a})(a+1)}{a-\sqrt{a}+1}\)
1) \(\sqrt{17-12\sqrt{2}}=\sqrt{9-12\sqrt{2}+8}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
=> a=3; b=2 => \(a^2+b^2=3^2+2^2=13\)
2) \(\dfrac{\sqrt{a}-1}{a\sqrt{a}+\sqrt{a}-a}:\dfrac{1}{a^2+a}=\dfrac{a\left(a-1\right)\left(a+1\right)}{\sqrt{a}\left(\sqrt{a}^3+1\right)}=\dfrac{\sqrt{a}\left(a^2-1\right)}{\sqrt{a}^3+1}\)
