2x + 23 = 2012 - ( 2012 - 15 )

2x + 23 = 2012 - 2012 + 15

2x + 23 = 15

2x = -8

x = -4

2x + 23 = 2012 - (2012 - 15)

2x + 23 = 2012 - 2012 + 15

2x + 23 = 0 + 15

2x + 23 = 15

2x         = -8

  x         = -4

\(2x+23=2012-\left(2012-15\right)\)

\(2x+23=2012-1997\)

\(2x+23=15\)

\(2x=15-23\)

\(2x=-8\)

\(x=-8:2\)

\(x=-4\)

\(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\left(2x-1\right)^{2012}-\left(2x-1\right)^{2010}=0\)

\(\Leftrightarrow[\left(2x-1\right)^{2010}.\left(2x-1\right)^2]-\left(2x-1\right)^{2010}=0\)\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1\right)^2-1]=0\)

\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-1-1\right)\left(2x-1+1\right)]=0\)

\(\Leftrightarrow\left(2x-1\right)^{2010}.[\left(2x-2\right)2x]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^{2010}\\2x\left(2x-2\right)=0\end{matrix}\right.=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x=0\\2x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=1\end{matrix}\right.\)

Vậy x \(\in\left\{\dfrac{1}{2};0;1\right\}\)

\((2x-1)^{2012} = (2x-1)^{2010} \)

\(​​\)\(\Leftrightarrow\)\((2x-1)^{2012} - (2x-1)^{2010} = 0\)

\(\Leftrightarrow\)\((2x-1)^{2010} . [(2x-1)^{2} - 1] = 0\)

\(\Leftrightarrow\)\((2x-1)^{2010} . (2x-2).2x = 0\)

\(\Leftrightarrow\)\(4 . (2x-1)^{2010} . (x-1) . x = 0\)

\(\Rightarrow\)\(\left[{}\begin{matrix}\left(2x-1\right)^{2010}=0\\x-1=0\\x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)

\(Vậy \) \(x= \)\(\dfrac{1}{2}\); \(x=1\) \(hay\) \(x=0\)

Xét vế trái biểu thức, ta có:
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot x\)
\(=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\right]\cdot x\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x\)
Xét vế phải biểu thức, ta có:
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
Từ đầu bài và 2 kết luận trên, ta suy ra:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)

X=2012

15 số chẵn = 30 số tự nhiên

Số lớn là: (2012 + 30) : 2 = 1021

Số bé là: 2012 - 1021 = 991

so lon 1021

so be 991

tk nhe@@@@@@@@@@@

bye$$