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\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{100!}< 2\)

\(2^1.2^2.2^3.....2^x=1024\Rightarrow2^{1+2+3+...+x}=2^{10}\)
\(\Rightarrow1+2+3+...+x=1024\Rightarrow x=4\)

\(1.\sqrt{x-1}=2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
\(2.\sqrt{3-x}=1\)
\(\Rightarrow3-x=1\)
\(\Rightarrow x=2\)
\(3.\left|x-1\right|+\left|x^2-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left|x^2-1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(4.\left|2x-3\right|-\left|x-1\right|=0\)
\(\Rightarrow\left|2x-3\right|=\left|x-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=x-1\\2x-3=-x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=3-1\\2x+x=3+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right..\)

X^3+9x=0
=>x(x^2-9)=0
=>x(x-3)(x+3)=0
=>x=0
Hoặc x-3=0 =>x=-3
Hoặc x+3=0 =>x=3
Vậy x=0,x=3;x=-3

a) Vì x< 0 nên x= \(-\sqrt{7}\)
b) x-2 =\(\sqrt{2}\)hoặc x-2 = -\(\sqrt{2}\)
suy ra x= \(\sqrt{2}\)+2 hoặc x= \(-\sqrt{2}\)+2
c)
x+\(\sqrt{3}\) =\(\sqrt{5}\)hoặc x+\(\sqrt{3}\) = -\(\sqrt{5}\)
suy ra x= \(\sqrt{5}-\sqrt{3}\)hoặc x= \(-\sqrt{5}-\sqrt{3}\)
Các bạn tự kết luận nhé

Ta có :\(A=3+3^2+3^3+...+3^{2008}\)(1)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)(2)
Lấy (2) trừ đi 1 ta có :
\(\Rightarrow2A=3^{2009}-3\)
Ta lại có :
\(2A+3=3^x\)
\(\Rightarrow3^{2009}=3^x\)
\(\Rightarrow x=2009\)
2\(^{x-3}\) - 3.2\(x\) + 92 = 0
2\(^{x-3}\) - 3.2\(^{x-3}\).23 + 92 = 0
2\(^{x-3}\).(1 - 3.23) + 92 = 0
2\(^{x-3}\).(1 - 24) + 92 = 0
2\(^{x-3}\).(-23) + 92 = 0
2\(^{x-3}\).(-23) = - 92
2\(^{x-3}\) = - 92 : (-23)
2\(^{x-3}\) = 4
2\(^{x-3}\) = 22
\(x-3\) = 2
\(x=2+3\)
\(x=5\)
Vậy \(x=5\)