a) \(\dfrac{2^6\cdot5^5}{10^5}\)

\(=\dfrac{2^6\cdot5^5}{2^5\cdot5^5}\)

\(=2\)

b) \(\dfrac{3^6\cdot5^7}{15^6}\)

\(=\dfrac{3^6\cdot5^7}{3^6\cdot5^6}\)

\(=5\)

a: 15 mod 2=1

b: 27 div 5=5

c: 17 mod 3=2

d: 21 div 2=10

e: 10 mod 4=2

f: 23 div 5=4

a) 25.2³ − 3² + 125 = 25.8 – 9 + 125 = 316.

b) 2.3² + 5.(2 + 3) = 2.9+5.5 = 43.

a) \(A=1-8x-x^2=-\left(x^2+8x+16\right)+17=-\left(x-4\right)^2+17\le17\)

\(ĐTXR\Leftrightarrow x=4\)

b) \(B=5-2x+x^2=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

\(ĐTXR\Leftrightarrow x=1\)

c) \(C=x^2+4y^2-6x+8y-2021=\left(x^2-6y+9\right)+\left(4y^2+8y+4\right)-2034=\left(x-3\right)^2+\left(2y+2\right)^2-2034\ge-2034\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

a: Ta có: \(A=-x^2-8x+1\)

\(=-\left(x^2+8x-1\right)\)

\(=-\left(x^2+8x+16-17\right)\)

\(=-\left(x+4\right)^2+17\le17\forall x\)

Dấu '=' xảy ra khi x=-4

b: Ta có: \(x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

\(A=\dfrac{45^{10}.5^{10}}{75^{10}}=\dfrac{5^{10}.9^{10}.5^{10}}{25^{10}.3^{10}}=\dfrac{5^{20}.3^{20}}{3^{10}.5^{20}}=3^{10}=59049\)

a, đặt \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

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

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

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

\(b,\)

\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)

\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)

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

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

\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)

a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

=3-1=2

b) Ta có: \(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

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

\(=4\sqrt{7}\)

a) Tại x=-1

\(\Rightarrow x^5-5=\left(-1\right)^5-5=-6\)

\(a,\)Thay \(x=-1\) vào \(x^5-5\)

\(\Rightarrow\left(-1\right)^5-5=-6\)

\(b,\) 

+ TH1:

Thay \(x=1\) vào \(x^2-3x-5\)

\(\Rightarrow1^2-3.1-5=-7\)

+TH2:

Thay \(x=-1\) vào \(x^2-3x-5\)

\(\Rightarrow\left(-1\right)^2-3.\left(-1\right)-5=-1\)

b) Tại x=1

\(\Rightarrow x^2-3x-5=1-3-5=-7\)

Tại x=-1

\(\Rightarrow x^2-3x-5=1+3-5=-1\)

a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)

b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)

\(B=10+5\sqrt{3}+10-5\sqrt{3}=20\)

a)

$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$

$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$

b)

$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$

$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$