\(\begin{array}{l}{\left( {a + b} \right)^3} = \left( {a + b} \right){\left( {a + b} \right)^2}\\\;\;\;\;\;\;\;\;\;\;\; = \left( {a + b} \right)\left( {{a^2} + 2ab + {b^2}} \right)\\\;\;\;\;\;\;\;\;\;\;\; = a.{a^2} + a.2ab + a.{b^2} + b.{a^2} + b.2ab + b.{b^2}\\\;\;\;\;\;\;\;\;\;\;\; = {a^3} + 2{a^2}b + a{b^2} + {a^2}b + 2a{b^2} + {b^3}\\\;\;\;\;\;\;\;\;\;\;\; = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\end{array}\)        \(\begin{array}{l}{\left( {a - b} \right)^3} = \left( {a - b} \right){\left( {a - b} \right)^2}\\\;\;\;\;\;\;\;\;\;\;\; = \left( {a - b} \right)\left( {{a^2} - 2ab + {b^2}} \right)\\\;\;\;\;\;\;\;\;\;\;\; = a.{a^2} - a.2ab + a.{b^2} - b.{a^2} + b.2ab - b.{b^2}\\\;\;\;\;\;\;\;\;\;\;\; = {a^3} - 2{a^2}b + a{b^2} - {a^2}b + 2a{b^2} - {b^3}\\\;\;\;\;\;\;\;\;\;\;\; = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\end{array}\)

\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)

a, Điều kiện xác định: \(\frac{1}{2}x + \frac{\pi }{4} \ne k\pi  \Leftrightarrow x \ne  - \frac{\pi }{2} + k2\pi ,k \in \mathbb{Z}.\)

Ta có: \(cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) =  - 1 \Leftrightarrow cot\left( {\frac{1}{2}x + \frac{\pi }{4}} \right) = \cot \left( { - \frac{\pi }{4}} \right)\)

\( \Leftrightarrow \frac{1}{2}x + \frac{\pi }{4} =  - \frac{\pi }{4} + k\pi  \Leftrightarrow x =  - \pi  + k2\pi ,k \in \mathbb{Z}\,\,(TM).\)

Vậy \(x =  - \pi  + k2\pi ,k \in \mathbb{Z}\,\).

b, Điều kiện xác định: \(3x \ne k\pi  \Leftrightarrow x \ne k\frac{\pi }{3},k \in \mathbb{Z}.\)

\(\;cot3x =  - \frac{{\sqrt 3 }}{3} \Leftrightarrow cot3x = \cot \left( { - \frac{\pi }{3}} \right)\)

\( \Leftrightarrow 3x =  - \frac{\pi }{3} + k\pi  \Leftrightarrow x =  - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\,(TM).\)

Vậy \(x =  - \frac{\pi }{9} + k\frac{\pi }{3},k \in \mathbb{Z}\,\).

\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)

Do \(0\le a,b,c\le1\) nên

\(\left(a^2-1\right)\left(b-1\right)\ge0\Leftrightarrow a^2b+1\ge a^2+b\)

Tương tự

\(3+a^2b+b^2c+c^2a\ge a^2+b^2+c^2+a+b+c\ge2\left(a^3+b^3+c^3\right)\)...

\(\begin{array}{l}a){\left( {\frac{2}{3}} \right)^{10}}{.3^{10}} = \frac{{{2^{10}}}}{{{3^{10}}}}{.3^{10}} = {2^{10}}\\b){( - 125)^3}:{25^3} = {( - 125:25)^3} = {( - 5)^3} =  - 125\\c){(0,08)^3}{.10^6} = {(0,08)^3}{.100^3} = {(0,08.100)^3} = {8^3}\end{array}\)

\(\begin{array}{l}a){( - 2)^3}.{( - 2)^4} = {( - 2)^{3 + 4}} = {( - 2)^7}\\b){(0,25)^7}:{(0,25)^3} = {(0,25)^{7 - 3}} = {(0,25)^4}\end{array}\)

a) Đặt: x = a- b; y = b - c ; z = c- a 

Ta có: x + y + z = 0 

=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)

=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

b) Đặt: \(a=x^2-2x\) 

Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)

\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)

d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)

Đặt: \(x^2-8=t\)

Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)

\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)

\(=\left(2x^2+9x-16\right)^2\)

\(\begin{array}{l}a)A = 32,125 - (6,325 + 12,125) - (37 + 13,675)\\ = 32,125 - 6,325 - 12,125 - 37 - 13,675\\ = (32,125 - 12,125) + ( - 6,325 - 13,675) - 37\\ = 20 + ( - 20) - 37\\ =  - 37\\b)B = 4,75 + {\left( {\frac{{ - 1}}{2}} \right)^3} + 0,{5^2} - 3.\frac{{ - 3}}{8}\\ = 4,75 + \frac{{ - 1}}{8} + 0,25 + \frac{9}{8}\\ = (4,75 + 0,25) + \left( {\frac{{ - 1}}{8} + \frac{9}{8}} \right)\\ = 5 + \frac{8}{8}\\ = 5 + 1\\ = 6\\c)C = 2021,2345.2020,1234 + 2021,2345.( - 2020,1234)\\ = 2021,2345.[2020,1234 + ( - 2020,1234)]\\ = 2021,2345.0\\ = 0\end{array}\)