có kết quả r mà

4/7 *1/3+5/7:2/3

\(\frac{4}{21}\)+ \(\frac{15}{14}\)= \(\frac{53}{42}\)

=4/21+15/14

=53/42

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

Ta có: \(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)

\(=\sqrt{\sqrt{7}^2-2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}+\sqrt{\sqrt{7}^2+2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(\left|2x-1\right|=\dfrac{3}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{3}{2}\\2x-1=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Thay \(x=\dfrac{5}{4}\) vào D ta có:

\(D=4x+3=4.\dfrac{5}{4}+3=5+3=8\)

Thay \(x=-\dfrac{1}{4}\) vào D ta có:

\(D=4.\dfrac{-1}{4}+3=-1+3=2\)

Để \(D=\dfrac{3}{2}\)

\(\Leftrightarrow4x+3=\dfrac{3}{2}\\ \Leftrightarrow4x=-\dfrac{3}{2}\\ \Leftrightarrow x=-\dfrac{3}{8}\)

\(\dfrac{6\times8\times11}{66\times4}=\dfrac{6\times4\times2\times11}{6\times11\times4}=2\)

A

Để biểu thức đề bài cho có giá trị nguyên thì \(5\sqrt{x}-6⋮2\sqrt{x}-3\)

\(\Leftrightarrow10\sqrt{x}-12⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\in\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow2\sqrt{x}\in\left\{0;2;4;6\right\}\)

hay \(x\in\left\{0;1;4;9\right\}\)