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Ta có : 8 . 6 + 288 : (x - 3)2 = 50
=> 48 + 288 : (x - 3)2 = 50
=> 288 : (x - 3)2 = 50 - 48
=> 288 : (x - 3)2 = 2
=> (x - 3)2 = 288 : 2
=> (x - 3)2 = 144
=> không tìm được x thỏa mãn điều kiện trên
8.6+288:(x-3)2 = 50
<=> 48+288:(x-3)2 = 50
<=> 288:(x-3)2 = 50-48 = 2
<=> (x-3)2 = 2.288 = 576 = 242
<=> \(\orbr{\begin{cases}x-3=24\\x-3=-24\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24+3=27\\x=3-24=-21\end{cases}}\)
Vậy x thuộc { 27 ; -21 }
\(6^2.2^2.5:\left[3.12-\left(2x-6\right)\right]=2^3.5\)
\(\Rightarrow36.4.5:\left[36-\left(2x-6\right)\right]=8.5\)
\(\Rightarrow720:\left[36-\left(2x-6\right)\right]=40\)
\(\Rightarrow36-\left(2x-6\right)=720:40=18\)
\(\Rightarrow2x-6=36-18=18\)
\(\Rightarrow2x=18+6=24\)
\(\Rightarrow x=24:2=12\)
8.6+288:(x-3)^2=50
48+288:(x-3)^2=50
288:(x-3)^2=50-48
288:(x-3)^2=2
(x-3)^2=288:2
(x-3)^2=144
(x-3)^2=12^2
x-3=12
x=12+3
x=15
Vậy x=15
HT
\(8\cdot6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48=2\)
\(\left(x-3\right)^2=288:2=144\)
Vì x là số tự nhiên nên (x-3) là số tự nhiên.
\(\left(x-3\right)^2=12^2\)
\(x-3=12\)
\(x=12+3=15\)
\(\Leftrightarrow\dfrac{3\cdot2^{x+1}}{2^{x+3}}=40\\ \Leftrightarrow\dfrac{3}{2^2}=40\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
62.22.x-1=288
(2.3)2.22.x-1=288
22.32.22.x-1=288
(22.22).32.x-1=288
(24.32).x-1=288
144.x-1=288
144.x=288+1
144.x=289
x=289:144
x=289/144