a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)

 \({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)

Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);

b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)

\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)

Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)

c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)

\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)

Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).

d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)

Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).

(-2) ^4 . (-2) 65 và ( -2) ^ 12 : ( -2) ^3

=( -2) ^ 4+5 =(-2)^9 và (-2) ^12-3 = ( -2) ^9 

vậy ( -2) ^9 = (-2) ^9 

Nên (-2) ^4 .( -2) ^5 = ( -2) ^ 12 : ( -2) ^3

a) \(\left( { + 4} \right).\left( { - 8} \right)\) là tích của hai số nguyên khác dấu nên mang dấu âm. Vậy \(\left( { + 4} \right).\left( { - 8} \right) < 0\)

b) \(\left( { - 3} \right).4\) là tích của hai số nguyên khác dấu nên mang dấu âm. Vậy\(\left( { - 3} \right).4 < 4\)

c) \(\left( { - 5} \right).\left( { - 8} \right)\) là tích của hai số nguyên âm nên \(\left( { - 5} \right).\left( { - 8} \right) = 5.8\)

\(\left( { + 5} \right).\left( { + 8} \right)\) là tích của hai số nguyên dương nên \(\left( { + 5} \right).\left( { + 8} \right) = 5.8\)

Vậy \(\left( { - 5} \right).\left( { - 8} \right) = \left( { + 5} \right).\left( { + 8} \right)\).

c.

(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})

Mà:

\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)

\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)

\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)

Lời giải:

a.

$5+\sqrt{2}>5+\sqrt{1}=6$

$4+\sqrt{3}< 4+\sqrt{4}=6$

$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$

b.

$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$

$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$

Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$

b: \(\sqrt{3}-1=\sqrt{4-2\sqrt{3}}\)

mà \(4-3\sqrt{3}< 4-2\sqrt{3}\)

nên \(\sqrt{4-3\sqrt{3}}< \sqrt{3}-1\)

Đề này sai rồi bạn vì \(4-3\sqrt{3}< 0\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

Ta có:

a) \(\sin \left( {x + 2\pi } \right) = \sin x\) với mọi \(x\; \in \;\mathbb{R}\)

b) \(\cos \left( {x + 2\pi } \right) = \cos x\) với mọi \(x\; \in \;\mathbb{R}\)

c) \(\tan \left( {x + \pi } \right) = \tan x\) với mọi \(x \ne \frac{\pi }{2} + k\pi ,\;k\; \in \;\mathbb{Z}\)

d) \(\cot \left( {x + \pi } \right) = \cot x\) với mọi \(x \ne \frac{\pi }{2} + k\pi ,\;k\; \in \;\mathbb{Z}\)

a) Vì \(\dfrac{1}{24}< \dfrac{1}{83}\) 

⇒ \(\dfrac{1}{24^9}>\dfrac{1}{83^{13}}\)

a) \(\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{27}\right)^9=\dfrac{1}{3^{27}}\)

\(\left(\dfrac{1}{83}\right)^{13}< \left(\dfrac{1}{81}\right)^{13}=\dfrac{1}{3^{52}}\)

Mà \(\dfrac{1}{3^{27}}>\dfrac{1}{3^{52}}\)

\(\Rightarrow\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)

b) \(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\)

Mà \(25^{100}< 27^{100}\)

\(\Rightarrow5^{199}< 3^{300}\)

\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)

\(a,\left(\dfrac{1}{24}\right)^9=\dfrac{1}{24^9};\left(\dfrac{1}{83}\right)^{13}=\dfrac{1}{83^{13}};24^9< 83^{13}\left(24< 83;9< 13\right)\\ \Rightarrow\dfrac{1}{24^9}< \dfrac{1}{83^{13}}\Rightarrow\left(\dfrac{1}{24}\right)^9< \left(\dfrac{1}{83}\right)^{13}\\ b,3^{300}=27^{100}>25^{100}=5^{200}>5^{199}\\ \Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)

a.

\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)

c.

\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)

d.

\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)

Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì

\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)

Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)

Vậy \(\left(a;b\right)=\left(9;4\right)\)

a) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\\ = \left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right] + \left[ {{{4.2}^3}.\left( {\sqrt 2 } \right) + 4.2.{{\left( {\sqrt 2 } \right)}^3}} \right]\\ = 68 + 48\sqrt 2 \end{array}\)

b) Áp dụng công thức nhị thức Newton, ta có

          \({\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\)

          \({\left( {2 - \sqrt 2 } \right)^4} = \left( {2 +(- \sqrt 2 )} \right)^4= {2^4} + {4.2^3}.\left( { - \sqrt 2 } \right) + {6.2^2}.{\left( { - \sqrt 2 } \right)^2} + 4.2.{\left( { - \sqrt 2 } \right)^3} + {\left( { - \sqrt 2 } \right)^4}\)

Từ đó,

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4} = 2\left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right]\\ = 2\left( {16 + 48 + 4} \right) = 136\end{array}\)

c) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {1 - \sqrt 3 } \right)^5} = \left( {1 +(- \sqrt 3 )} \right)^5=  1 + 5.\left( { - \sqrt 3 } \right) + 10.{\left( { - \sqrt 3 } \right)^2} + 10.{\left( { - \sqrt 3 } \right)^3} + 5.{\left( { - \sqrt 3 } \right)^4} + 1.{\left( { - \sqrt 3 } \right)^5}\\ = \left[ {1 + 10.{{\left( { - \sqrt 3 } \right)}^2} + 5.{{\left( { - \sqrt 3 } \right)}^4}} \right] + \left[ {5.\left( { - \sqrt 3 } \right) + 10.{{\left( { - \sqrt 3 } \right)}^3} + 1.{{\left( { - \sqrt 3 } \right)}^5}} \right]\\ = 76 - 44\sqrt 3 \end{array}\)