Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

Giá của chiếc ti vi sau lần giảm thứ nhất là:

    17 820 000 : (100% -10%) = 19 800 000 (đồng)

Giá của chiếc ti vi lúc ban đầu là: 

    19 800 000 : (100% - 10%) = 22 000 000 (đồng)

Kết luận giá của chiếc ti vi ban đầu là: 22 000 000 đồng

giúp tui

\(\dfrac{9}{14}-\dfrac{5}{8}+\dfrac{3}{7}=\dfrac{36}{56}-\dfrac{35}{56}+\dfrac{24}{56}=\dfrac{25}{56}\)

\(\dfrac{2}{9}-\dfrac{7}{10}+\dfrac{4}{5}=\dfrac{20}{90}-\dfrac{63}{90}+\dfrac{72}{90}=\dfrac{29}{90}\)

\(\dfrac{8}{19}+\dfrac{7}{5}-\dfrac{9}{20}+\dfrac{11}{12}=\dfrac{8}{19}+\dfrac{84}{60}-\dfrac{27}{60}+\dfrac{55}{60}=\dfrac{8}{19}+\dfrac{112}{60}=\dfrac{8}{19}+\dfrac{28}{15}=\dfrac{652}{285}\)

giúp tui với

\(-\dfrac{11}{12}+\dfrac{7}{8}+\dfrac{1}{10}\)

\(=-\dfrac{1}{24}+\dfrac{1}{10}\)

\(=\dfrac{7}{120}\)

a) \(\left|a+b\right|\le\left|a\right|+\left|b\right|\)
b) \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\) với \(\left|a\right|\ge\left|b\right|\)
c) \(\left|ab\right|\le\left|a\right|.\left|b\right|\)
d) \(\left|\dfrac{a}{b}\right|\le\dfrac{\left|a\right|}{\left|b\right|}\)

A = \(x^2\) - 20\(x\) + 2024

A = (\(x^2\) - 20\(x\) + 100) + 1924

A = (\(x\) - 10)² + 1924

Vì ( \(x-10\))² ≥ 0 ⇒ (\(x-10\))² + 1924 ≥ 1924

     Vậy A_min  =  1924 ⇔ \(x\) = 10 

Kết luận giá trị nhỏ nhất của A là 1924 xảy ra khi \(x\) = 10

\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)

\(=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{7}\right)+...+\left(\dfrac{1}{512}+\dfrac{1}{513}+...+\dfrac{1}{1023}\right)+\dfrac{1}{1024}< 1+\dfrac{1}{2}.2+\dfrac{1}{2^2}.2^2+...+\dfrac{1}{2^9}.2^9+\dfrac{1}{1024}\)

\(=1+1+1+1+...+\dfrac{1}{1024}=10+\dfrac{1}{1024}< 11\left(đpcm\right)\)

5 + | \(x\) + 3| = 9

      |\(x+3\)| = 9 - 5

      |\(x\) + 3| = 4

      \(\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -7; 1}

A = 2² - 20\(x\) + 2024 

A = 4 + 2024  - 20\(x\)

A = 2028 - 20\(x\) 

Ta có đồ thị 

Không có giá trị nhỏ nhất em nhá