a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

a)>

b)>

c)>

tầm là thế

\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)

\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)

Vì 3⁶⁹ > 3⁶⁴

\(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)

\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)

\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)

Vì 3⁶⁹ > 3⁶⁴

=> \(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)

=> \(\left(\frac{1}{27}\right)^{23}< \left(\frac{1}{81}\right)^{16}\)

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)

a) \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)

\(=\frac{15}{6}-\frac{1}{6}.9+\frac{3}{8}\)

\(=\frac{15}{6}-\frac{9}{6}+\frac{3}{8}\)

\(=1+\frac{3}{8}\)

\(=\frac{11}{8}\)

b) \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=2^2.3=12\)

a/ \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)

= \(\frac{15}{6}-\frac{3}{18}.9+\frac{8}{8}\)

= \(\frac{15}{6}-\frac{3}{2}+\frac{3}{8}\)

= \(\frac{60-36+9}{24}=\frac{33}{24}=\frac{11}{8}\)

b/ \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}\) \(=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=\frac{2^2.3^{35}}{3^{34}}=\frac{4.3}{1}=12\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

   Bạn để ý: 81 = 3^4 (3⁴)            27 = 3^3

            64 = 2^6                256 = 2^8

 Vậy \(\left(\frac{27}{64}\right)^{15}=\left(\frac{3^2}{8^2}\right)^{15}=\left(\frac{3}{8}\right)^{30};\left(\frac{81}{256}\right)^{10}=\left(\frac{3^4}{4^4}\right)^{10}=\left(\frac{3}{4}\right)^{40}\)

Vì 3/8 <3/4 ; 30<40 nên \(\left(\frac{3}{8}\right)^{30}

a) x=1

b) x=1

c) x= -(245/81)

d) x= 1/27

e) x=3

g) x=4

cần gấp

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)

=> \(A=\frac{121}{243}\)

=> \(2x+\frac{12}{25}=\frac{121}{243}\)

=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)

=> x = ......