Ý bạn là sao nhỉ? 

Theo mình hiểu thì bạn muốn biến 72 thành căn đúng không? Vậy thì bạn chỉ cần biểu diễn $72=\sqrt{72^2}=\sqrt{5184}$ thôi.

À. Khi em muốn biến đổi $\sqrt{x}$ để đưa thừa số ra dấu căn, em cần phân tích $x$ dưới dạng $x=a.b^2$ với $a$ là nhận giá trị $1$ hoặc $a$ không phải số chính phương có thể.

Ví dụ: 

$72$ thì em phân tích thành $2.6^2$. Khi đó $\sqrt{72}=\sqrt{2.6^2}=\sqrt{2}.\sqrt{6^2}=6\sqrt{2}$

$81$ thì em phân tích thành $9^2$ từ đó $\sqrt{81}=\sqrt{9^2}=9$

..........

Bạn chỉ cần phân tích nó ra thành thừa số nguyên tố là xong

\(75\%:2\dfrac{1}{5}+\left(0,5\right)^2.\left(-7\right)+2,5\left(7\dfrac{2}{3}+5\dfrac{2}{3}\right)\)

\(=\dfrac{3}{4}:\dfrac{11}{5}+\dfrac{1}{4}.\left(-7\right)+2,5\left(12+\dfrac{4}{3}\right)\)

\(=\dfrac{15}{44}-\dfrac{7}{4}+\dfrac{5}{2}.\dfrac{40}{3}\)

\(=-\dfrac{31}{22}+\dfrac{100}{3}\)

\(=\dfrac{2107}{66}\)

\(\dfrac{0.375-0.3+\dfrac{3}{11}+\dfrac{3}{12}}{-0.625+0.5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1.5+1-0.75}{2.5+\dfrac{5}{3}-1.25}\)

=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)

=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)

=\(\dfrac{3}{-5}+\dfrac{3}{5}\)

=\(0\)

Sau số 2,7 là một đẳng thức nhé

a) Ta có: \(\left|2.5-x\right|=1.3\)

\(\Leftrightarrow\left|x-2.5\right|=1.3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)

Vậy: \(x\in\left\{3.8;1.2\right\}\)

b) Ta có: \(1.6-\left|x-0.2\right|=0\)

\(\Leftrightarrow\left|x-0.2\right|=1.6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)

Vậy: \(x\in\left\{1.8;-1.4\right\}\)

c) Ta có: \(13^x=169\)

\(\Leftrightarrow13^x=13^2\)

\(\Leftrightarrow x=2\)

Vậy: x=2

d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)

\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)

\(\Leftrightarrow x^2=\dfrac{16}{25}\)

hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

\(|2,5-x|=1,3\)

\(\Rightarrow2,5-x=1,3\)    hoặc      -(2,5-x)=1,3

=>x=2,5-1,3                          x-2,5=1,3

=>x=1,2                                x=1,3+2,5=3,8

Vậy \(x\in\left\{1,2;3,8\right\}\)

\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)

=> x-0,2=1,6       hoặc       -(x-0,2)=1,6

=>x=1,6+0,2                      0,2-x=1,6

=>x=3,8                              x=0,2-1,6=-1,4

Vậy \(x\in\left\{3,8;-1,4\right\}\)

\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)

\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)

\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)

\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)

\(C=\dfrac{1\dfrac{1}{2}+0,6-\dfrac{3}{7}}{2,5+1-\dfrac{35}{49}}=\dfrac{\dfrac{3}{2}+\dfrac{3}{5}-\dfrac{3}{7}}{\dfrac{5}{2}+\dfrac{5}{5}-\dfrac{5}{7}}=\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{7}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{7}\right)}=\dfrac{3}{5}\)

3/5