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\(\dfrac{\left(\dfrac{2}{5}\cdot\sqrt{16}+2\cdot\sqrt{\dfrac{16}{24}}\right)}{2}\cdot\sqrt{\dfrac{1}{16}}\)
=\(\dfrac{\left(\dfrac{2}{5}\cdot4+2\cdot\sqrt{\dfrac{2}{3}}\right)}{2}\cdot\dfrac{1}{4}\)
=\(\dfrac{\left(\dfrac{8}{5}+\dfrac{2\cdot\sqrt{6}}{3}\right)}{8}\)
=\(\dfrac{\left(\dfrac{24}{15}+\dfrac{5\cdot\left(2\cdot\sqrt{6}\right)}{15}\right)}{8}\)
=\(\dfrac{\left(\dfrac{24+10\cdot\sqrt{6}}{15}\right)}{8}\)
=\(\dfrac{2\cdot\left(12+5\cdot\sqrt{6}\right)}{120}\)
=\(\dfrac{12+5\cdot\sqrt{6}}{60}\)
a) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)
\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
2)
a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72
\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)
\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)
Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)
Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)
b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12
\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)
\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)
Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)
Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)
\(d,x-5\sqrt{x}=0\)
\(ĐKXĐ:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}\)(Thỏa mãn ĐKXĐ)
Vậy...
\(\sqrt{\dfrac{16}{169}}.\dfrac{-3}{2}.\left(\dfrac{3}{2}+\dfrac{-5}{12}\right):\left(-\dfrac{1}{2}\right)\\ =\left|\sqrt{\left(\pm\dfrac{4}{13}\right)^2}\right|.\dfrac{-3}{2}.\dfrac{13}{12}.\left(-2\right)\\ =\left(\dfrac{4}{13}.\dfrac{13}{12}\right).\left(-2.\dfrac{-3}{2}\right)\\ =\dfrac{4}{12}.3=\dfrac{12}{12}=1\)
\(\left(1,5+2\dfrac{1}{2}-2\sqrt{2^2}\right):\left(4\dfrac{1}{4}-\sqrt{1,96}+0,9\right)\)
= ( 1,5 + 2,5 - 2.2 ) : (4,25 - 1,4 + 0,9)
= ( 4 - 4 ) : ( 3,75 )
= 0 : 3,75
= 0
Chúc bạn học tốt
a)
ĐKXĐ: \(2x\geq 0\Leftrightarrow x\geq 0\)
Vậy TXĐ của $x$ là \(D= [0;+\infty)\)
b)
ĐK: \((2x-1)(x+3)\neq 0\Leftrightarrow \left\{\begin{matrix} 2x-1\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow \Leftrightarrow \left\{\begin{matrix} x\neq \frac{1}{2}\\ x\neq -3\end{matrix}\right.\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{1}{2}; -3\right\}\)
c)
ĐK: \(8x^3+1\neq 0\Leftrightarrow x^3\neq \frac{-1}{8}\Leftrightarrow x\neq \frac{-1}{2}\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{-1}{2}\right\}\)
d)
ĐK:
\(|x-2015|+1\neq 0\Leftrightarrow |x-2015|\neq -1\Leftrightarrow x\in\mathbb{R}\)
Vậy TXĐ \(D=\mathbb{R}\)
e)
ĐK: \(\left\{\begin{matrix} |x-1,2|\neq 0\\ 2x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 1,2\\ x\neq 2,5\end{matrix}\right.\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{1,2; 2,5\right\}\)
f)
ĐK: \(x^2-4\neq 0\Leftrightarrow (x-2)(x+2)\neq 0\Leftrightarrow x\neq \pm 2\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{\pm 2\right\}\)
Ta có \(6=\sqrt{36}\)
\(\sqrt{37}-\sqrt{14}=\sqrt{37}+\left(-\sqrt{14}\right)\)
\(6-\sqrt{15}=\sqrt{36}-\sqrt{15}=\sqrt{36}+\left(-\sqrt{15}\right)\)
Vì \(\sqrt{37}>\sqrt{36}\) và \(-\sqrt{14}>-\sqrt{15}\)
\(\Rightarrow\sqrt{37}+\left(-\sqrt{14}\right)>\sqrt{36}+\left(-\sqrt{15}\right)\)
\(\Rightarrow\sqrt{37}-\sqrt{14}>\sqrt{36}-\sqrt{15}\)
hay \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
Chúc bn học tốt 
\(\sqrt{\dfrac{1}{9}}\cdot\sqrt{0.81}+\sqrt{0.09}\)
=\(\dfrac{1}{3}\cdot\dfrac{3}{10}+\dfrac{3}{10}\)
=\(\dfrac{1}{10}+\dfrac{3}{10}\)
=\(\dfrac{2}{5}\)