a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)

⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)

⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{1}{18}\)

⇒ x + 1 = 18

⇒ x = 17

Vậy x = 17

b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)

⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)

⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)

⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)

⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)

⇒ \(\frac{1}{x+3}=\frac{1}{148}\)

⇒ x + 3 = 148

⇒ x = 145

Vậy x = 145

Giải:

1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)

\(=7^8.\left(\dfrac{1}{7}\right)^8\)

\(=7^8.\dfrac{1^8}{7^8}\)

\(=1\)

2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)

\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)

\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)

\(=1\)

3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)

\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)

\(=1\)

4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)

\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)

\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)

\(=\dfrac{5}{13}\)

Vậy ...

@Hắc Hường dạo này lên hình lại rồi ak:))

P1 có 2 thừa số âm nhân với nhau nên P1 >0

P2 có 3 thừa số âm nhân với nhau nên P2 <0

P3 vì trong dấu ba chấm có thừa số 0/10 nên P3 =0

Vậy P2 < P3 < P1

KO khó lắm đâu. Mong bạn hiểu để bài sau tương tự thì làm được.

CHúc bạn học tốt.

a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)

\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)

\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)

b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)

\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)

\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)

c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)

\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)

\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)

\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180

\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)

\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)

Tính nốt vì số dữ quá , lần sau để số ít thôi

\(-\frac{2000}{139}\)

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)