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2 . 3x + 5 . 3x+1 = 153
3x. ( 2 + 5.3 ) = 153
3x . 17 = 153
3x = 9 = 32
x = 2
Vậy x = 2
\(2.3^x+5.3^{x+1}=153\)
\(\Leftrightarrow2.3^x+5.3^x.3=153\)
\(\Leftrightarrow3^x.\left(2+5.3\right)=153\)
\(\Leftrightarrow3^x.17=153\)
\(\Leftrightarrow3^x=153:17\)
\(\Leftrightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)
\(\Rightarrow x=2\)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
\(2.3^x=10.3^{12}+8.27^4\\ \Rightarrow2.3^x=10.3^{12}+8.\left(3^3\right)^4\\ \Rightarrow2.3^x=10.3^{12}+8.3^{12}\\ \Rightarrow2.3^x=3^{12}\left(10+8\right)\\ \Rightarrow2.3^x=3^{12}.18\)
\(=>3^x=3^{12}.18:2\\ \Rightarrow3^x=3^{12}.3^2\\ \Rightarrow3^x=3^{10}\)
Ta có: 2.3x+1-3x=135
3x+1=135
3x=134
3x=Vô lý
Bn xem lại đề bài đi bn
2.3 mũ x.3-3 mũ x=135
3 mũx.(2+3)=135
3mũx.5=135
3mũx=135:5
3 mũ x=3 mũ 3