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a: \(3x-\left|2x+1\right|=2\)
\(\Leftrightarrow\left|2x+1\right|=3x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2\right)^2-\left(2x+1\right)^2=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2-2x-1\right)\left(3x-2+2x+1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(5x-1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow x=3\)
e: Ta có: \(2n-3⋮n+1\)
\(\Leftrightarrow2n+2-5⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{0;-2;4;-6\right\}\)
7^6+7^5+7^4 chia hết cho 11
= 7^4.2^2+7^4.7+7^4
= 7^4.(2^2+7+1)
= 7^4. 11
Vì tích này có số 11 nên => chia hết cho 7
1. \(A=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)-\left(\frac{2}{5}-\frac{5}{7}+\frac{4}{35}\right)+\frac{1}{41}\)
\(=\left(\frac{5}{6}+\frac{1}{6}\right)-\left(\frac{-11}{35}+\frac{4}{35}\right)+\frac{1}{41}\)\(=1-\frac{-7}{35}+\frac{1}{41}=1+\frac{1}{5}+\frac{1}{41}=\frac{251}{205}\)
2. a) \(1+4+4^2+4^3+......+4^{99}=\left(1+4\right)+\left(4^2+4^3\right)+.......+\left(4^{98}+4^{99}\right)\)
\(=\left(1+4\right)+4^2\left(1+4\right)+.........+4^{98}\left(1+4\right)\)
\(=5+4^2.5+........+4^{98}.5=5\left(1+4^2+.....+4^{98}\right)⋮5\)( đpcm )
b) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1+1}.5=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\)( đpcm )