1,S=1-2+3-4+...+25-26

  S=[1-2]+[3-4]+...+[25-26]

  S=[-1]+[-1]+...+[-1]

  S=[-1]*13

  S= -13

2,S=1-3+5-7+...-53+55

  S=[1-3]+[5-7]+...+[51-53]+55

  S=[-2]+[-2]+...+[-2]+55

  S=[-2]*27+55

  S=[-54]+55

  S=1

Gọi số cần tìm là \(\overline{a0b}\)( \(a;b\inℕ\)và \(a\ne0\)và \(0\le a;b\le9\))

Theo đề bài, ta có:

\(\overline{a0b}=6.\overline{ab}\)\(\Leftrightarrow100a+b=6.\left(10a+b\right)\)

                          \(\Leftrightarrow100a+b=60a+6b\)

                           \(\Rightarrow100a-60a=6b-b\)

                          \(\Leftrightarrow40a=5b\)

                           \(\Rightarrow\frac{a}{b}=\frac{5}{40}=\frac{1}{8}\)

Mà \(a;b\inℕ\)và \(a\ne0\)và \(0\le a;b\le9\)

\(\Rightarrow a=1;b=8\)

Vậy số cần tìm là 108

128-3(x+4)=23

3(x+4)=128-23

3(x+4)=105

x+4=105:3

x+4=35

x=35-4

x=31

Vậy x=31

Học tốt

ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0

khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)

*ĐK : 101x\(\ge\) 0 => x\(\ge\)0

=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)

     \(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)

...

\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

Ta có :

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)

<=> \(100x+\frac{5050}{101}=101x\)

<=> \(100x-101x=\frac{-5050}{101}\)

<=> -x = -50

=> x = 50 ( t/m x >/ 0)

=>3-2x =0 =>\(\orbr{\begin{cases}2x=3\\2x=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=1.5\\x=-1.5\end{cases}}\)

=>4y+5=0=>4y=-5=>y=-1.25                                                                                                                                                                                Vậy x=1.5 hoặc x=-1.5 , y=-1.25

\((x+1)\):\((x+2)\)\(\varepsilon\) \(ℤ\)

suy ra: \((x+1)\)-\((x+2)\) chia hết cho \((x+2)\)

suy ra: \((-1)\) chia hết cho\((x+2)\)

suy ra:\((x+2)\) \(\varepsilon\)Ư\((-1)\)

SUY RA: X=\((-1)\)-2=-3

Vậy x=[-3]

2008.2072007-2007.20082008=1.00638062

hocj toot

2008.2072007-2007.20082008=1.00638062