Question

Bài 1

A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}\) + \(\dfrac{1}{\sqrt{2}+\sqrt{3}}\) + \(\dfrac{1}{\sqrt{3}+\sqrt{4}}\) + ......+ \(\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

Bài 2:tìm x,biết

\(\dfrac{\sqrt{3x-1}}{\sqrt{5x+2}}=7\)

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Answer 1

Bài 1:

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

\(=-\sqrt{1}+\sqrt{100}=100-1=99\)

Bài 2:

\(\dfrac{\sqrt{3x-1}}{\sqrt{5x+2}}=7\)

\(pt\Leftrightarrow\sqrt{\dfrac{3x-1}{5x+2}}=7\)\(\Leftrightarrow\dfrac{3x-1}{5x+2}=49\)

\(\Leftrightarrow3x-1=49\left(5x+2\right)\)

\(\Leftrightarrow3x-1-245x-98=0\)

\(\Leftrightarrow-242x-99=0\Rightarrow x=-\dfrac{9}{22}\)