\(B=3^1+3^2+3^3+...+3^{100}\\3B=3^2+3^3+3^4+...+3^{101}\\3B-B=(3^2+3^3+3^4+...+3^{101})-(3^1+3^2+3^3+...+3^{100})\\2B=3^{101}-3\\\Rightarrow 2B+3=3^{101}\)

Mặt khác: \(2B+3=3^n\)

\(\Rightarrow 3^n=3^{101}\\\Rightarrow n=101(tm)\)

Vậy n = 101.

cảm ơn bạn nha :))

Ta có: \(a+b=5\Rightarrow a=5-b\)

Thay \(a=5-b\) vào \(2a-b=4\) ta có:

\(2\cdot\left(5-b\right)-b\)

\(\Rightarrow10-2b-b=4\)

\(\Rightarrow10-3b=4\)

\(\Rightarrow3b=10-4\)

\(\Rightarrow3b=6\)

\(\Rightarrow b=\dfrac{6}{3}=2\)

Lúc này ta tìm được \(a\):

\(a=5-b=5-2=3\)

Vậy: \(a=3,b=2\)

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\(4,\\ 2.B=\sqrt{x}-1+\dfrac{2-2\sqrt{x}}{\sqrt{x}}\left(x>0\right)\\ B=\dfrac{x-\sqrt{x}+2-2\sqrt{x}}{\sqrt{x}}=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}\)

\(3.x=\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\left(3+\sqrt{2}\right)+\left(3-\sqrt{2}\right)=6\)

Thay vào B, ta được \(B=\dfrac{6-3\sqrt{6}+2}{\sqrt{6}}=\dfrac{6\sqrt{6}-18+2\sqrt{6}}{6}=\dfrac{4\sqrt{6}-9}{3}\)

\(4.B=0\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}=0\Leftrightarrow x-3\sqrt{x}+2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

\(7.B\in Z\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}-3+\dfrac{2}{\sqrt{x}}\in Z\\ \Leftrightarrow\dfrac{2}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{1;4\right\}\left(\sqrt{x}>0\right)\)

+ Nếu n chia hết cho 3 thì tích chia hết cho 3

+ Nếu n chia 3 dư 1 thì 2n chia 3 dư 2 => 2n+1 chia hết cho 3 => tích chia hết cho 3

+ nếu n chia 3 dư 2 => n+1 chia hết cho 3 => tích chia hết cho 3

=> tích chia hết cho 3 với mọi n

c) \(\left(\dfrac{1}{2}\cdot x+\dfrac{1}{4}\right)\cdot\left(2x-\dfrac{1}{3}\right)=0\)

 \(\dfrac{1}{2}\cdot x+\dfrac{1}{4}=0\)

     \(\dfrac{1}{2}\cdot x=0-\dfrac{1}{4}\)

     \(\dfrac{1}{2}\cdot x=-\dfrac{1}{4}\)

          \(x=-\dfrac{1}{4}\div\dfrac{1}{2}\)

          \(x=-\dfrac{1}{2}\)

\(2x-\dfrac{1}{3}=0\)

\(2x=0+\dfrac{1}{3}\)

\(2x=\dfrac{1}{3}\)

  \(x=\dfrac{1}{3}\div2\)

  \(x=\dfrac{1}{6}\)

\(\Rightarrow\) \(x=\) {\(-\dfrac{1}{2};\dfrac{1}{6}\)}

a) Ta có: \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)

\(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-\sqrt{3}+5\cdot\dfrac{2}{\sqrt{3}}\)

\(=-9\sqrt{3}+\dfrac{10}{\sqrt{3}}\)

\(=\dfrac{-27+10}{\sqrt{3}}=\dfrac{-17\sqrt{3}}{3}\)

2: Thay x=1 và y=-4 vào (d), ta được:

2m+2=-4

hay m=-3