\(\left(-\dfrac{3}{4}\right)3x-1=\dfrac{256}{81}\)

\(-\dfrac{9}{4}\cdot x=\dfrac{337}{81}\)

\(x=\dfrac{337}{81}\cdot-\dfrac{4}{9}\)

\(x=-\dfrac{1348}{729}\)

 \(a.5^{27}=\left(5^3\right)^9=125^9\\ 2^{63}=\left(2^7\right)^9=128^9\)

Vì 128⁹ > 125⁹ => 2⁶³ > 5²⁷

^{\(5^{28}=\left(5^4\right)^7=625^7\\ 2^{63}=\left(2^9\right)^7=512^7\)}

Vì 625⁷ > 512⁷ = > 5²⁸ > 2⁶³

Đã CMR: \(5^{27}< 2^{63}< 5^{28}\)

\(b.A=1+2+2^2+2^3+2^4\\ 2A=2+2^2+2^3+2^4+2^5\\ 2A-A=\left(2+2^2+2^3+2^4+2^5\right)-\left(1+2+2^2+2^3+2^4+\right)\\ A=2^5-1\\ 2^5-1=2^5-1=>A=B\\ c,C=3+3^2+....+3^{100}\\ 3C=3^2+......+3^{101}\\ 3C-C=\left(3^2+...+3^{101}\right)-\left(3+...+3^{100}\right)\\ 2C=3^{101}-3\\ C=\dfrac{3^{101}-3}{2}\\ \dfrac{3^{101}-3}{2}=\dfrac{3^{101}-3}{2}=>C=D\)

\(A=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\times4}\)

\(A\cdot\dfrac{1}{7}=\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\)

\(A\cdot\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)

\(A\cdot\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{28}\)

\(A\cdot\dfrac{1}{7}=\dfrac{13}{28}\)

\(A=\dfrac{13}{4}\)

\(\dfrac{2^{11}\cdot51-2^{10}\cdot27}{2^9\cdot5^2}\)

\(=\dfrac{2^{10}\left(102-27\right)}{2^9\cdot5^2}\)

\(=\dfrac{2^{10}\cdot75}{2^9\cdot5^2}\)

\(=\dfrac{2^{10}\cdot5^2\cdot3}{2^9\cdot5^2}\)

\(=2\cdot3=6\)

Tổng diện tích 2 đáy là:

1610 x 2 : 20 = 161 (m)

Độ dài đáy bé là:

(161 - 14) : 2=  73,5 ( m)

Độ dài đáy lớn là:

(161 + 14) : 2 = 87,5 (m)

Đ/s:....

\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)

Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)

b,\(3^{500}=\left(3^5\right)^{100}=243^{100}\)

Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)

CMR: ASE+A1+E1=ACE

a, 1/6 + 1/3

= 1/6 + 2/6

= 3/6 

= 1/2

b, 1/4 + 1/8

= 2/8 + 1/8

= 3/8

c, -5/12 + 2/-3

= -5/12 - 2/3

= -5/12 - 8/12

= -13/12