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nhớ tick nha

Bài 1

a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{33}{99}-\frac{1}{99}\)

\(=\frac{32}{99}\)

c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Tk mình nha!!

Câu 2:

\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)

\(=\frac{3\cdot100}{2}\)

\(=\frac{300}{2}=150\)

\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)

x=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5

x=1-1/5

x=4/5

cau a dau nhi cuoi cung k phai j dau nha ! mk an lom ! 

\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)

 \(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)

\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)

ta có |x+5| \(\ge\)0 \(\forall x\)

Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn

b,

 \(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)

\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}:\frac{1}{3}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1504}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1504}\)

\(\frac{1}{x+3}=-\frac{11}{7502}\)

\(x+3=\left(7502.1\right):\left(-11\right)\)

\(x+3=7502:\left(-11\right)\)

\(x+3=-682\)

\(x=-682-3\)

\(x=-385\)

0000000000

 \(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\) 

=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\) 

=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)

2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)

\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\) 

\(\dfrac{35}{12}-\dfrac{7}{5}\)

\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)

4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)

(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\) 

6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)

7 + \(\dfrac{103}{28}\)

\(\dfrac{299}{28}\)

x. \(\dfrac{1}{2}\) =\(\dfrac{3}{4}+\dfrac{1}{5}\)

x . \(\dfrac{1}{2}=\dfrac{15}{20}+\dfrac{4}{20}\)

x.\(\dfrac{1}{2}=\dfrac{19}{20}\)

x      = \(\dfrac{19}{20}:\dfrac{1}{2}\)

x = 19/10

\(x-\dfrac{1}{5}=\dfrac{2}{3}\) . \(\dfrac{9}{4}\)

\(x-\dfrac{1}{5}=\) \(\dfrac{3}{2}\)

x = \(\dfrac{3}{2}+\dfrac{1}{5}\)

x =\(\dfrac{15}{10}\) + \(\dfrac{3}{10}\)

x=\(\dfrac{18}{10}\) =\(\dfrac{9}{5}\)