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sai r ban oi
( 1!x1 + 2!x2 + .... +19! x 19)= (2-1) x 1! + (3 - 1) x 2! + ...+ (20-1) x 19!
= 2! - 1! + 3! - 2! + ... + 20!- 19!
=-1! + 20!
21!-21= 20! x 21 - 21
=(20! - 1 )x 21
=> (20!-1) x21
20! - 1
=21
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)
\(\left(4+2^2+2^3+2^4+...+2^{10}\right)x=2^{22}-2^{21}\)
\(\Rightarrow\left(2^2+2^2+2^3+...+2^{10}\right)x=2^{21}\left(2-1\right)\)
\(\Rightarrow2^{20}.x=2^{21}\) (Vì \(2^2+2^2=2^3\))
\(\Rightarrow x=2\)
Vậy x=2
a, \(\left(-2\right)x-\left(-21\right)=15\)
\(=>\left(-2\right)x+21=15\)
\(=>\left(-2\right)x=15-21\)
\(=>\left(-2\right)x=-6\)
\(=>x=\left(-6\right):\left(-2\right)\)
\(=>x=3\)
________
b, \(\left(3x-2^2\right).7^3=7^4\)
\(=>3x-4=7^4:7^3\)
\(=>3x-4=7\)
\(=>3x=7+4\)
\(=>3x=11\)
\(=>x=\dfrac{11}{3}\)
ai giúp minh minh ****
2^2+2^3+2^4+...+2^6=2^21
X=6