Question

P=\(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+2}{1-a^2}\)

a rut gon P

b tìm a để P =\(\dfrac{7}{8}\)

Answer 1

Lời giải:

ĐK: \(a\geq 0; a\neq 1\)

Ta có:

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

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

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

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

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

b)

\(P=\frac{7}{8}\Leftrightarrow \frac{a^2-a+1}{a^2-1}=\frac{7}{8}\)

\(\Rightarrow 8(a^2-a+1)=7(a^2-1)\)

\(\Leftrightarrow a^2-8a+15=0\Rightarrow \left[\begin{matrix} a=5\\ a=3\end{matrix}\right.\) (đều thỏa mãn)