Question

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

1.So sánh giá trị của M với 1

2.Rút gọn

Answer 1

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

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

\(A=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(A=\frac{\sqrt{a}-1}{\sqrt{a}}\)

1. Ta có \(\sqrt{a}-1< \sqrt{a}\Leftrightarrow\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)

Vậy \(A< 1\)

Answer 2

ĐKXĐ tự viết nhé

2. Rút gọn

A=\(\left(\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

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

1. đề sai