M=3+3mũ2+3mũ3+3mũ4+……+3mũ2020
Tìm X€:biết 2×m+3=3mũ X
2.Tìm X biết:
A,720:[41-(2×y-5)=2mũ3 ×5
B,(y+1)+(y+2)+(y+3) +….+(y+100)=5705
M=3+3mũ2+3mũ3+3mũ4+……+3mũ2020
Tìm X€:biết 2×m+3=3mũ X
2.Tìm X biết:
A,720:[41-(2×y-5)=2mũ3 ×5
B,(y+1)+(y+2)+(y+3) +….+(y+100)=5705
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
M = 3 + {3^2} + {3^3} + {3^4} + …. + {3^{2020}}\\
\Leftrightarrow 3M = 3.\left( {3 + {3^2} + {3^3} + {3^4} + …. + {3^{2020}}} \right)\\
\Leftrightarrow 3M = 3.3 + {3.3^2} + {3.3^3} + {3.3^4} + ….. + {3.3^{2020}}\\
\Leftrightarrow 3M = {3^2} + {3^3} + {3^4} + {3^5} + …. + {3^{2021}}\\
\Rightarrow 3M – M = \left( {{3^2} + {3^3} + {3^4} + {3^5} + …. + {3^{2021}}} \right) – \left( {3 + {3^2} + {3^3} + {3^4} + …. + {3^{2020}}} \right)\\
\Leftrightarrow 2M = {3^{2021}} – 3\\
2M + 3 = {3^x}\\
\Leftrightarrow \left( {{3^{2021}} – 3} \right) + 3 = {3^x}\\
\Leftrightarrow {3^{2021}} = {3^x}\\
\Leftrightarrow x = 2021\\
2,\\
a,\\
720:\left[ {41 – \left( {2y – 5} \right)} \right] = {2^3}.5\\
\Leftrightarrow 720:\left[ {41 – \left( {2y – 5} \right)} \right] = 8.5\\
\Leftrightarrow 720:\left[ {41 – \left( {2y – 5} \right)} \right] = 40\\
\Leftrightarrow 41 – \left( {2y – 5} \right) = 720:40\\
\Leftrightarrow 41 – \left( {2y – 5} \right) = 18\\
\Leftrightarrow 2y – 5 = 41 – 18\\
\Leftrightarrow 2y – 5 = 23\\
\Leftrightarrow 2y = 23 + 5\\
\Leftrightarrow 2y = 28\\
\Leftrightarrow y = 28:2\\
\Leftrightarrow y = 14\\
b,\\
1 + 2 + 3 + …. + n = \dfrac{{n.\left( {n + 1} \right)}}{2}\\
\left( {y + 1} \right) + \left( {y + 2} \right) + \left( {y + 3} \right) + ….. + \left( {y + 100} \right) = 5705\\
\Leftrightarrow \left( {y + y + y + …. + y} \right) + \left( {1 + 2 + 3 + …. + 100} \right) = 5705\,\,\,\,\,\left( {100\,\,so\,\,y} \right)\\
\Leftrightarrow 100.y + \dfrac{{100.\left( {100 + 1} \right)}}{2} = 5705\\
\Leftrightarrow 100.y + \dfrac{{100.101}}{2} = 5705\\
\Leftrightarrow 100.y + 5050 = 5705\\
\Leftrightarrow 100.y = 5705 – 5050\\
\Leftrightarrow 100.y = 700\\
\Leftrightarrow y = 700:100\\
\Leftrightarrow y = 7
\end{array}\)