cái này chắc đề sai

(2 x X - 1)mũ 3 = (2 x X - 2)mũ 2

A=1/2x(1/2+1/6+1/12+...+1/10100)

  =1/2x(1/1x2+1/2x3+1/3x4+...+1/100x101)

  =1/2x(1-1/2+1/2-1/3+1/3+1/4+...+1/100-1/101)

  =1/2x(1-1/101)

 =1/2x100/101=50/101 (đúng thì cho mk nhoa)

a: \(A=\frac{1}{1\cdot1}+\frac{1}{2\cdot3}+\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\cdots+\frac{1}{50\cdot99}\)

\(=2\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\right)\)

\(=2\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left\lbrack\left(1+\frac12+\frac13+\frac14+\cdots+\frac{1}{100}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\right\rbrack\)

\(=2\left\lbrack1+\frac12+\frac13+\cdots+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\right\rbrack\)

\(=2\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)\)

b: Ta có: \(\frac{1}{51}>\frac{1}{75};\frac{1}{52}>\frac{1}{75};\ldots;\frac{1}{75}=\frac{1}{75}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (1)

Ta có: \(\frac{1}{76}>\frac{1}{100};\frac{1}{77}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}=\frac{25}{100}=\frac14\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)>\frac13+\frac14=\frac{7}{12}\)

=>\(A>\frac{7}{12}\cdot2=\frac76\)

1.

Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}$

$=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1$

$\Rightarrow a=2(-1)=-2; b=3(-1)=-3; c=4(-1)=-4$

2.

$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}$
$=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)

b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).

Vậy : \(S=\dfrac{99}{100}.\)

a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)

\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)

b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)