ĐKXĐ: x>=0; x<>9

\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

\(\left(3x-4\right)^3=5^2+4.5^2\)

\(\Leftrightarrow\left(3x-4\right)^3=5^2\left(1+4\right)\)

\(\Leftrightarrow\left(3x-4\right)^3=5^3\)

\(\Leftrightarrow3x-4=5\Leftrightarrow3x=9\Leftrightarrow x=3\)

Ta có: \(\left(3x-4\right)^3=5^2+4\cdot5^2\)

\(\Leftrightarrow3x-4=5\)

hay x=3

so sánh à

c: \(9^{15}=3^{30}\)

\(27^{10}=3^{30}\)

Do đó: \(9^{15}=27^{10}\)

d: \(2^{333}=8^{111}\)

\(3^{222}=9^{111}\)

Do đó: \(2^{333}< 3^{222}\)

Lời giải:

$A=7+(7^2+7^3+7^4+7^5)+(7^6+7^6+7^8+7^9)+....+(7^{2018}+7^{2019}+7^{2020}+7^{2021})$

$=7+7^2(1+7+7^2+7^3)+7^6(1+7+7^2+7^3)+....+7^{2018}(1+7+7^2+7^3)$

$=7+(1+7+7^2+7^3)(7^2+7^6+....+7^{2018}$

$=7+400(7^2+7^6+....+7^{2018})$

Dễ thấy $400(7^2+7^6+....+7^{2018})$ tận cùng là $0$ 

Do đó $A$ tận cùng là $7$

\(3\left(x+2\right)^3-1^{2019}=5\cdot4^2\)

\(\Leftrightarrow3\left(x+2\right)^3=5\cdot16+1=81\)

\(\Leftrightarrow x+2=3\)

hay x=1

\(4^{15}.9^{15}< 2^n.3^n< 18^{16}.2^{16}\)

⇒\(\left(4.9\right)^{15}< \left(2.3\right)^n< \left(18.2\right)^{16}\)

⇒\(\left(6^2\right)^{15}< 6^n< \left(6^2\right)^{16}\)

⇒\(6^{30}< 6^n< 6^{32}\)

⇒\(6^n=6^{31}\)

⇒n=31

\(4^{15}\cdot9^{15}< 2^n\cdot3^n< 18^{16}\cdot2^{16}\\ \Leftrightarrow\left(4\cdot9\right)^{15}< \left(2\cdot3\right)^n< \left(18\cdot2\right)^{16}\\ \Leftrightarrow36^{15}< 6^n< 36^{16}\\ \Leftrightarrow6^{30}< 6^n< 6^{32}\\ \Leftrightarrow n=31\)

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