10000 câu trắc nghiệm tổng hợp môn Toán 2025 mới nhất (có đáp án) - Phần 1

(‒0,25)⁴.4⁴

\[ = {\left( { - \frac{1}{4}} \right)^4} \cdot {4^4}\]

\[ = {\left( { - \frac{1}{4} \cdot 4} \right)^4}\]

\[ = {\left( { - \frac{4}{4}} \right)^4}\]

= (‒1)⁴ = 1.

\[\left( {1 + \frac{7}{9}} \right) \cdot \left( {1 + \frac{7}{{20}}} \right) \cdot \left( {1 + \frac{7}{{33}}} \right) \cdot .. \cdot \left( {1 + \frac{7}{{2900}}} \right)\]

\[ = \frac{{16}}{9} \cdot \frac{{27}}{{20}} \cdot \frac{{40}}{{33}} \cdot ... \cdot \frac{{2\,\,907}}{{2\,\,900}}\]

\[ = \frac{{2 \cdot 8 \cdot 3 \cdot 9 \cdot 4 \cdot 10 \cdot ... \cdot 51 \cdot 57}}{{1 \cdot 9 \cdot 2 \cdot 10 \cdot 3 \cdot 11 \cdot ... \cdot 50 \cdot 58}}\]

\[ = \frac{{\left( {2 \cdot 3 \cdot 4 \cdot \cdot \cdot 51} \right)\left( {8 \cdot 9 \cdot 10 \cdot \cdot \cdot 57} \right)}}{{\left( {1 \cdot 2 \cdot 3 \cdot \cdot \cdot 50} \right)\left( {9 \cdot 10 \cdot 11 \cdot \cdot \cdot 58} \right)}}\]

\[ = \frac{{51 \cdot 8}}{{50 \cdot 58}}\]

\[ = \frac{{51 \cdot 2 \cdot 2 \cdot 2}}{{25 \cdot 2 \cdot 2 \cdot 29}}\]

\[ = \frac{{102}}{{725}}.\]

Ta có:

\[\frac{{2013}}{1} + \frac{{2012}}{2} + \frac{{2011}}{3} + ... + \frac{2}{{2012}} + \frac{1}{{2013}}\]

\[ = \left( {1 + \frac{{2012}}{2}} \right) + \left( {1 + \frac{{2011}}{3}} \right) + ... + \left( {1 + \frac{2}{{2012}}} \right) + \left( {1 + \frac{1}{{2013}}} \right) + 1\]

\[ = \frac{{2014}}{2} + \frac{{2014}}{3} + ... + \frac{{2014}}{{2012}} + \frac{{2014}}{{2013}} + \frac{{2014}}{{2014}}\]

\[ = 2014 \cdot \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2012}} + \frac{1}{{2013}} + \frac{1}{{2014}}} \right)\]

Nên từ đề bài, ta có:

\[\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2012}} + \frac{1}{{2013}} + \frac{1}{{2014}}} \right) \cdot x = 2014 \cdot \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2012}} + \frac{1}{{2013}} + \frac{1}{{2014}}} \right)\]

Suy ra x = 2014.

(145 × 99 + 145) ‒ (143 × 102 ‒ 143)

= [(145 × (99 + 1)] ‒ [(143 × (102 ‒ 1)]

= (145 × 100) ‒ (143 × 101)

= 14500 ‒ 14443

= 57.

A = (‒2) + (‒59) ‒ (‒22) + 59

= ‒2 ‒ 59 + 22 + 59

= (‒2 + 22) + (‒59 + 59)

= 20.

\[\frac{{{2^{23}} + {\rm{ }}{2^{24}} + {\rm{ }}{2^{25}}}}{{{2^{18}} + {2^{19}} + {2^{20}}}}\]

\[ = \frac{{{2^{18}}\left( {{2^5} + {\rm{ }}{2^6} + {\rm{ }}{2^7}} \right)}}{{{2^{18}}\left( {{2^0} + {2^1} + {2^2}} \right)}}\]

\[ = \frac{{{2^5}\left( {{2^0} + {\rm{ }}{2^1} + {\rm{ }}{2^2}} \right)}}{{{2^0} + {2^1} + {2^2}}}\]

= 2⁵ = 32.

\[\frac{{{2^3} \cdot {9^4} + {9^3} \cdot 45}}{{{9^2} \cdot 10 - {9^2}}}\]

\[ = \frac{{8 \cdot {9^4} + {9^3} \cdot 9 \cdot 5}}{{{9^2} \cdot 10 - {9^2}}}\]

\[ = \frac{{8 \cdot {9^4} + {9^4} \cdot 5}}{{{9^2} \cdot 10 - {9^2} \cdot 1}}\]

\[ = \frac{{{9^4} \cdot \left( {8 + 5} \right)}}{{{9^2} \cdot \left( {10 - 1} \right)}}\]

\[ = \frac{{{9^4} \cdot 13}}{{{9^3}}}\]

= 9.13 = 117.

Lời giải:

(2x ‒ 1)³ = 27

(2x ‒ 1)³ = 3³

2x ‒ 1 = 3

2x = 4

x = 2

Vậy x = 2.