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\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\\ 3^x\cdot3^2+4\cdot3^x\cdot3=5103\\ 3^x\left(9+12\right)=5103\\ 3^x\cdot21=5103\\ 3^x=243\\ 3^x=3^5\\ x=5\)
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\(4\cdot3^{x+2}-3^{x-1}=963\\ \Leftrightarrow36\cdot3^x-\dfrac{1}{3}\cdot3^x=963\\ \Leftrightarrow3^x=27\\ \Leftrightarrow x=3\)
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\(\left(4.3x-4\right)-2=18\)
\(\Rightarrow12x-4-2=18\)
\(\Rightarrow12x=24\Leftrightarrow x=2\)
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\(2.3^x+5.3^x.3=153\Leftrightarrow2.3^x+15.3^x=153\)
\(\Leftrightarrow17.3^x=153\Leftrightarrow3^x=9\Leftrightarrow x=2\)
\(\Leftrightarrow2.3^x+5.3.3^x=153\)
\(\Leftrightarrow17.3^x=3^2.17\)
\(\Leftrightarrow3^x=3^2\Rightarrow x=2\)
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Nguyễn Thị Mai:
\(^{\frac{1}{2.3}x+\frac{1}{3.4}x+....+\frac{1}{49.50}x=1}\)
Đặt: \(^{\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}+0+0+...+0-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\frac{12}{25}x=1}\)
\(^{\Leftrightarrow x=\frac{25}{12}}\)
Tham khảo nha ~~
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a: =>3[(2x-1)^2-4]=49*125:175+196=231
=>(2x-1)^2-4=77
=>(2x-1)^2=81
=>2x-1=9 hoặc 2x-1=-9
=>x=5 hoặc x=-4
b: \(\Leftrightarrow2\cdot3^x\cdot3-4^3=7^2\cdot\left(27-25\right)\)
=>\(6\cdot3^x=49\cdot2+64=162\)
=>3^x=27
=>x=3
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Lời giải:
a.
$3[(2x-1)^2-4]-14^2=7^2.5^3:175=35$
$3[(2x-1)^2-4]=35+14^2=231$
$(2x-1)^2-4=231:3=77$
$(2x-1)^2=77+4=81=9^2=(-9)^2$
$\Rightarrow 2x-1=9$ hoặc $2x-1=-9$
$\Rightarrow x=5$ hoặc $x=-4$
b.
$2.3^{x+1}-4^{10}:4^7=(7^5:7^3).(3^3-5^2)=7^2.2=98$
$2.3^{x+1}-4^3=98$
$2.3^{x+1}=98+4^3=162$
$3^{x+1}=162:2=81=3^4$
$\Rightarrow x+1=4$
$\Rightarrow x=3$
Ta có: \(4.3^{x+2}-2.3^{x+1}=810\)
\(4.3^{x+1}.3-2.3^{x+1}=810\)
\(2.3^{x+1}\left(2.3-1\right)=810\)
\(2.3^{x+1}.5=810\)
\(3^{x+1}=81=3^4\)
x+1=4
x=3