\(a.\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)
a, Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=-2\)
b, \(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\) (ĐK: x ≠ -2, x ≠ 2)
\(\Leftrightarrow\dfrac{10-\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2-4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow10-x-2=x^2-4\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
Vậy...
\(b.\)
\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\left(x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\cdot\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=1\)
\(\Leftrightarrow10-\left(x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow8-x-x^2+4=0\)
\(\Leftrightarrow-x^2-x+12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)
