Thứ tự thực hiện các phép tính. Quy tắc chuyển vế

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

2x + 3 = -x + 6

2x + x = 6 – 3

3x = 3

x = 1

Vậy x = 1

\(\begin{array}{l} - 2x + {\left( { - \frac{2}{5}} \right)^2} = 0,{1^2}\\ \Leftrightarrow  - 2x + \frac{4}{{25}} = \frac{1}{{100}}\\ \Leftrightarrow  - 2x = \frac{1}{{100}} - \frac{4}{{25}}\\ \Leftrightarrow  - 2x = \frac{1}{{100}} - \frac{{16}}{{100}}\\ \Leftrightarrow  - 2x = \frac{{ - 15}}{{100}}\\ \Leftrightarrow x = \frac{{ - 15}}{{100}}:( - 2)\\ \Leftrightarrow x = \frac{{ - 15}}{{100}}.\frac{{ - 1}}{2}\\ \Leftrightarrow x = \frac{3}{{40}}\end{array}\)

\(\frac{{{{25}^{30}}}}{{{{125}^{15}}}} = \frac{{{{25}^{30}}}}{{{{(5.25)}^{15}}}} = \frac{{{{25}^{30}}}}{{{5^{15}}{{.25}^{15}}}} = \frac{{{{25}^{15}}}}{{{5^{15}}}} = {5^{15}}\)

\(\begin{array}{l}T = \left[ {\left( { - 43,57} \right).40 - 40.56,43} \right]:\left[ { - {7^2}.53,6 - 49.46,4} \right]\\T = 40.\left( { - 43,57 - 56,43} \right):\left[ { - 49.53,6 - 49.46,4} \right]\\T = \left[ {40.\left( { - 100} \right)} \right]:\left[ {49\left( {53,6 - 46,4} \right)} \right]\\T = \dfrac{{40.\left( { - 100} \right)}}{{49.\left( { - 100} \right)}} = \dfrac{{40}}{{49}}\end{array}\)

\(\left( { - 2x + \frac{5}{2}} \right).\left( {{x^2} + 4} \right) = 0\)

+) Trường hợp 1:

\(\begin{array}{l} - 2x + \frac{5}{2} = 0\\ \Leftrightarrow 2x = \frac{5}{2}\\ \Leftrightarrow x = \frac{5}{2}:2\\ \Leftrightarrow x = \frac{5}{4}\end{array}\)

+) Trường hợp 2:

x² + 4 = 0

\( \Leftrightarrow {x^2} =  - 4\) ( Vô lí vì x² \( \ge \)0 với mọi x)

Vậy x = \(\frac{5}{4}\)

Q = 3ⁿ⁺³ + 3ⁿ⁺¹ + 2ⁿ⁺² + 2ⁿ⁺¹

= 3ⁿ⁺¹ . 3² + 3ⁿ⁺¹ + 2ⁿ⁺¹ . 2 + 2ⁿ⁺¹

= 3ⁿ⁺¹ . (3² + 1) + 2ⁿ⁺¹ . (2 + 1)

= 3ⁿ⁺¹ . 10 + 2ⁿ⁺¹ . 3

= 3ⁿ⁺¹ . 2.5 + 2ⁿ⁺¹ . 3

= 3.2 . ( 3ⁿ . 5 + 2)

= 6. ( 3ⁿ . 5 + 2)

Vì 6\( \vdots \) 6 nên 6. ( 3ⁿ . 5 + 2) \( \vdots \) 6 với mọi n nguyên dương

Vậy Q luôn chia hết cho 6

\(\begin{array}{l}\dfrac{{{8^7} + {8^7} + {8^7} + {8^7}}}{{{3^7} + {3^7} + {3^7}}}:\dfrac{{{2^7} + {2^7}}}{{{6^7} + {6^7} + {6^7} + {6^7} + {6^7} + {6^7}}} = {2^n}\\ \Leftrightarrow \dfrac{{{{4.8}^7}}}{{{{3.3}^7}}}:\dfrac{{{{2.2}^7}}}{{{{6.6}^7}}} = {2^n}\\ \Leftrightarrow \dfrac{{{{4.8}^7}}}{{{3^8}}}:\dfrac{{{2^8}}}{{{6^8}}} = {2^n}\\ \Leftrightarrow \dfrac{{{{4.8}^7}}}{{{3^8}}}.\dfrac{{{6^8}}}{{{2^8}}} = {2^n}\\ \Leftrightarrow \dfrac{{{2^2}.{{({2^3})}^7}{{.6}^8}}}{{{{(3.2)}^8}}} = {2^n}\\ \Leftrightarrow \dfrac{{{2^2}{{.2}^{21}}{{.6}^8}}}{{{6^8}}} = {2^n}\\ \Leftrightarrow {2^{23}} = {2^n}\\ \Leftrightarrow 23 = n\end{array}\)

Vậy n = 23

\(\begin{array}{l}B = 1,2.(3\frac{1}{3} - 2,2) - \frac{2}{{15}}.( - 2 + \frac{5}{6}) - {2022^0}\\ = \frac{{12}}{{10}}.(\frac{{10}}{3} - \frac{{11}}{5}) - \frac{2}{{15}}.(\frac{{ - 12}}{6} + \frac{5}{6}) - 1\\ = \frac{6}{5}.(\frac{{50}}{{15}} - \frac{{33}}{{15}}) - \frac{2}{{15}}.(\frac{{ - 7}}{6}) - 1\\ = \frac{6}{5}.\frac{{17}}{{15}} + \frac{7}{{45}} - 1\\ = \frac{{34}}{{25}} + \frac{7}{{45}} - 1\\ = \frac{{306}}{{225}} + \frac{{35}}{{225}} - \frac{{225}}{{225}}\\ = \frac{{116}}{{225}}\end{array}\)

Vì (2x+1)⁴ \( \ge \) 0, với mọi x nên (2x+1)⁴ +2 \( \ge \) 2, với mọi x

\( \Rightarrow \frac{3}{{{{(2x + 1)}^4} + 2}} \le \frac{3}{2}\), với mọi x. Dấu “=” xảy ra khi 2x + 1 = 0 hay x = \(\frac{{ - 1}}{2}\)

Vậy Max M = \(\frac{3}{2}\).