Question

Cho biểu thức M = \(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)

a) rút gọn M

b) Tính giá trị M nếu \(\dfrac{a}{b}=\dfrac{3}{2}\)

c) Tìm điều kiện của a , b để M < 1

Answer 1

a) \(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}.\dfrac{a-\sqrt{a^2-b^2}}{b}\)
\(=\dfrac{ab-\left(\sqrt{a^2-b^2}.a-a^2+b^2+a^2-a\sqrt{a^2-b^2}\right)}{b.\sqrt{a^2-b^2}}\)
\(=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=\sqrt{\dfrac{a-b}{a+b}}\)
b) \(M\) khi \(\dfrac{a}{b}=\dfrac{3}{2}=>2a=3b=>a=\dfrac{3b}{2}=>a-b=\dfrac{3b}{2}-b=\dfrac{1}{2}b\)
\(a+b=\dfrac{3b}{2}+b=\dfrac{5b}{2}\)
\(=>\dfrac{a-b}{a+b}=\dfrac{\dfrac{1}{2}b}{\dfrac{5}{2}b}=\dfrac{1}{2}.\dfrac{2}{5}=\dfrac{1}{5}=>M=\sqrt{\dfrac{1}{5}}\)
c) \(M< 1=>0\le1=>\dfrac{a-b}{a+b}-1\le0\)
\(< =>\dfrac{a-b-\left(a+b\right)}{a+b}\le0< =>\dfrac{-2b}{a+b}\le0\)