Question

Rút gọn:

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

Tính A với \(a=3+2\sqrt{2}\)

Answer 1

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a-1}\right)\cdot\left(\sqrt{a}+1\right)}\right)\)

Answer 2

.. tiếp

\(=\dfrac{a-1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

...

Answer 3

- Rút gọn:

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\dfrac{a-1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{1}{\sqrt{a}-1}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)

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

\(=\dfrac{a-1}{\sqrt{a}}\)

- Tính A với \(a=3+2\sqrt{2}\)

Thay \(a=3+2\sqrt{2}\) vào biểu thức \(A=\dfrac{a-1}{\sqrt{a}}\)

Ta có: \(\dfrac{3+2\sqrt{2}-1}{3+2\sqrt{2}}=\dfrac{2+2\sqrt{2}}{3+2\sqrt{2}}\)

\(=\left(2+2\sqrt{2}\right)\cdot\left(3-2\sqrt{2}\right)\)

\(=6-4\sqrt{2}+6\sqrt{2}-8\)

\(=-2+2\sqrt{2}\)

Vậy giá trị biểu thức A khi \(a=3+2\sqrt{2}\) là \(-2+2\sqrt{2}\)