Bài 3: Liên hệ giữa phép nhân và phép khai phương

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=\sqrt{3.4}+2\sqrt{3.9}+3\sqrt{5.5.3}-9\sqrt{4.4.3}\)

\(=2\sqrt{3}+2.3\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(\left(2\sqrt{2}-\sqrt{3}\right)^2=8-2\cdot2\sqrt{2}\cdot\sqrt{3}+3=11-4\sqrt{6}\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-\sqrt{2}^2=1+2\sqrt{3}+3-2=2+2\sqrt{3}\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-2=4+2\sqrt{3}-2=2+2\sqrt{3}\)

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}=6+2\sqrt{9-5}=6+2.2=10\)

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\dfrac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)

\(\sqrt{\dfrac{2x-3}{x-1}}=2\) ( đk: \(x\ne1\))

\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

\(\Leftrightarrow4\left(x-1\right)=2x-3\)

\(\Leftrightarrow4x-4=2x-3\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\) (tm)

Vậy...

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

\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x-3-4\left(x-1\right)=0\)

\(\Leftrightarrow2x-3-4x+4=0\)

\(\Leftrightarrow-2x+1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

ĐKXĐ: \(x\ne1\)

=>\(\dfrac{2x-3}{x-1}=4\)

<=>\(2x-3=4x-4\)

<=>\(2x-1=0\)

<=>\(x=\dfrac{1}{2}\)(nhận)

điều kiện : \(\left\{{}\begin{matrix}x\ge\dfrac{-3}{2}\\\left[{}\begin{matrix}x\le\dfrac{-3}{2}\\x\ge\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

ta có : \(\sqrt{4x^2-9}=2\sqrt{2x+3}\Leftrightarrow4x^2-9=4\left(2x+3\right)\)

\(\Leftrightarrow4x^2-14x+6x-21=0\Leftrightarrow2x\left(2x-7\right)+3\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

vậy \(x=\dfrac{-3}{2}\overset{.}{,}x=\dfrac{7}{2}\)