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\(A=2+2^2+2^3+...+2^{98}+2^{99}+2^{100}\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{97}.\left(1+2\right)+2^{99}.\left(1+2\right)\)
\(\Rightarrow A⋮3\)
\(\left[\left(3x+1\right)^3\right]^5=15^0\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1^{15}\)
\(\Rightarrow3x+1=1\)
\(\Leftrightarrow3x=1-1\)
\(\Leftrightarrow3x=0\Rightarrow x=0\)
\(\left[(3\times+1)^3\right]^5=15^0\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1^5\)
\(\Rightarrow(3\times+1)^3=1\)
\(\Rightarrow(3\times+1)^3=1^3\)
\(\Rightarrow3\times+1=1\)
\(\Rightarrow3\times=1-1\)
\(\Rightarrow3\times=0\)
\(\Rightarrow\times=0\)
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=3+2^2.\left(1+2+4\right)+...+2^{98}.\left(1+2+4\right)\)
\(=3+7.\left(2^2+2^5+...+2^{98}\right)\)chia 7 dư 3
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(S=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{98}+2^{99}+2^{100}\right)\)
\(S=\left(1+2+4\right)+2^3\left(1+2+4\right)+.....+2^{98}\left(1+2+4\right)\)
\(S=7+2^3\cdot7+....+2^{98}\cdot7\)
\(S=7\left(1+2^3+...+2^{98}\right)\)
=> S chia 7 dư 0 hay S chia hết cho 7
375:32-(38:36-2.23)
= 375 : 9 - ( 9 - 16 )
= \(\frac{125}{3}-9+16\)
= \(\frac{146}{3}\)