\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{25}{100}-\dfrac{10}{100}+\dfrac{1}{100}\\ =\dfrac{16}{100} =\dfrac{4}{25}\)

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{4}{25}\)

\(a.\dfrac{2}{3}-\left(-\dfrac{1}{2}-x\right)=-\dfrac{4}{5}\\ \dfrac{2}{3}+\dfrac{1}{2}+x=-\dfrac{4}{5}\\ x=-\dfrac{4}{5}-\dfrac{2}{3}-\dfrac{1}{2}\\ x=-\dfrac{59}{30}\\ b.\left(-x-3\dfrac{1}{4}\right)-\left(1\dfrac{2}{3}-2\dfrac{3}{4}\right)=\dfrac{-5}{6}\\ \left(-x-\dfrac{13}{4}\right)-\left(\dfrac{5}{3}-\dfrac{11}{4}\right)=\dfrac{-5}{6}\\ -x-\dfrac{13}{4}-\dfrac{5}{3}+\dfrac{11}{4}=-\dfrac{5}{6}\\ -x-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{4}{3}\\ c.\dfrac{8}{23}\cdot\dfrac{46}{24}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ d.\dfrac{x-1}{16}=\dfrac{3}{x+1}\\ \left(x-1\right)\left(x+1\right)=3\cdot16=48\\ x^2-1=48\\ x^2=49\\ x^2=7^2\\ x=\pm7\)

\(e.\left(1,2\right)^3x^2=\left(1,2\right)^5\\ x^2=\dfrac{\left(1,2\right)^5}{\left(1,2\right)^3}\\ x^2=\left(1,2\right)^2\\ x=\pm1,2\\ f.\left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=4\\ \left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=2^2\\TH1:\dfrac{2}{3}x-\dfrac{1}{4}=2\\ \dfrac{2}{3}x=2+\dfrac{1}{4}=\dfrac{9}{4}\\ x=\dfrac{9}{4}:\dfrac{2}{3}=\dfrac{27}{8}\\ TH2:\dfrac{2}{3}x-\dfrac{1}{4}=-2\\ \dfrac{2}{3}x=-2+\dfrac{1}{4}=-\dfrac{7}{4}\\ x=\dfrac{-7}{4}:\dfrac{2}{3}=-\dfrac{21}{8}\\ g.\left(\dfrac{1}{6}x-3\right)^2=\dfrac{4}{9}\\ \left(\dfrac{1}{6}x-3\right)^2=\left(\dfrac{2}{3}\right)^2\\ TH1:\dfrac{1}{6}x-3=\dfrac{2}{3}\\ \dfrac{1}{6}x=\dfrac{2}{3}+3=\dfrac{11}{3}\\ x=\dfrac{11}{3}:\dfrac{1}{6}=22\\ TH2:\dfrac{1}{6}x-3=-\dfrac{2}{3}\\ \dfrac{1}{6}x=-\dfrac{2}{3}+3=\dfrac{7}{3}\\ x=\dfrac{7}{3}:\dfrac{1}{6}=14\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(B=\dfrac{x-1}{x+1}-\dfrac{x-1}{x+1}-\dfrac{4}{1-x^2}\)

\(=-\dfrac{4}{1-x^2}=\dfrac{4}{x^2-1}\)

\(x^2-x=0\)

=>x(x-1)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Khi x=0 thì \(B=\dfrac{4}{0^2-1}=\dfrac{4}{-1}=-4\)

c: B=-3

=>\(\dfrac{4}{x^2-1}=-3\)

=>\(x^2-1=-\dfrac{4}{3}\)

=>\(x^2=-\dfrac{4}{3}+1=-\dfrac{1}{3}< 0\)

=>Không có giá trị nào của x thỏa mãn

d: Để B nguyên thì \(4⋮x^2-1\)

=>\(x^2-1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x^2\in\left\{2;0;3;5\right\}\)

mà x nguyên

nên x=0

e: Để B<0 thì \(\dfrac{4}{x^2-1}< 0\)

=>\(x^2-1< 0\)

=>\(x^2< 1\)

=>-1<x<1

mà x nguyên

nên x=0

f: Để B>=0 thì \(\dfrac{4}{x^2-1}>=0\)

=>x²-1>0

=>(x-1)(x+1)>0

=>\(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

\(a.\dfrac{1}{2}-3x=-\dfrac{2}{5}\\ 3x=\dfrac{1}{2}+\dfrac{2}{5}\\ 3x=\dfrac{9}{10}\\ x=\dfrac{9}{10}:3\\ x=\dfrac{3}{10}\\ b.-x+\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{1}{2}+\dfrac{5}{6}\\ x=\dfrac{4}{3}\\ c.x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\\ x+\dfrac{3}{5}=\dfrac{4}{25}\\ x=\dfrac{4}{25}-\dfrac{3}{5}\\ x=-\dfrac{11}{25}\\ d.\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\\ x=\dfrac{1}{7}:-\dfrac{3}{14}=-\dfrac{2}{3}\\ e.-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\\ \dfrac{1}{7}-x=\dfrac{1}{21}:-\dfrac{1}{3}=-\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\\ h.\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\\ \dfrac{1}{4}-3x+\dfrac{3}{2}=-\dfrac{3}{4}\\ 3x=\dfrac{1}{4}+\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{5}{2}\\ x=\dfrac{5}{2}:3\\ x=\dfrac{5}{6}\\ i.\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\\ \dfrac{2}{3}+2x=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\\ 2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{23}{21}\\ x=\dfrac{-23}{21}:2=-\dfrac{23}{42}\)

a: \(\dfrac{1}{2}-3x=-\dfrac{2}{5}\)

=>\(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{5}{10}+\dfrac{4}{10}=\dfrac{9}{10}\)

=>\(x=\dfrac{9}{10}:3=\dfrac{9}{30}=\dfrac{3}{10}\)

b: \(-x+\dfrac{1}{2}=-\dfrac{5}{6}\)

=>\(-x=-\dfrac{5}{6}-\dfrac{1}{2}=-\dfrac{5}{6}-\dfrac{3}{6}=-\dfrac{8}{6}=-\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}\)

c: \(x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\)

=>\(x+\dfrac{3}{5}=\dfrac{4}{25}\)

=>\(x=\dfrac{4}{25}-\dfrac{3}{5}=\dfrac{4}{25}-\dfrac{15}{25}=-\dfrac{11}{25}\)

d: \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)

=>\(\dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\)

=>\(x=-\dfrac{1}{7}:\dfrac{3}{14}=-\dfrac{1}{7}\cdot\dfrac{14}{3}=-\dfrac{2}{3}\)

e: \(-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\)

=>\(\dfrac{1}{7}-x=\dfrac{1}{21}:\dfrac{-1}{3}=\dfrac{-1}{21}\cdot3=-\dfrac{1}{7}\)

=>\(x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\)

h: \(\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\)

=>\(-3x+\dfrac{5}{4}=-\dfrac{3}{4}\)

=>\(-3x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)

=>\(x=\dfrac{-2}{-3}=\dfrac{2}{3}\)

i: \(\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\)

=>\(2x+\dfrac{2}{3}=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\)

=>\(2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{9}{21}-\dfrac{14}{21}=-\dfrac{23}{21}\)

=>\(x=-\dfrac{23}{21}:2=-\dfrac{23}{42}\)

cái gì dợ

trình và trinh

\(a.\dfrac{3}{7}\cdot\dfrac{5}{8}+\dfrac{3}{7}\cdot\dfrac{11}{8}+\dfrac{11}{7}=\dfrac{3}{7}\cdot\left(\dfrac{5}{8}+\dfrac{11}{8}\right)+\dfrac{11}{7}=\dfrac{3}{7}\cdot2+\dfrac{11}{7}=\dfrac{6}{7}+\dfrac{11}{7}=\dfrac{17}{7}\\ b.\dfrac{3}{8}\cdot19\dfrac{1}{3}-\dfrac{3}{8}\cdot\left(33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot-14=\dfrac{-21}{8}\\ c.\dfrac{1}{3}\cdot\dfrac{5}{4}+\dfrac{1}{3}\cdot\dfrac{7}{4}-2022^0=\dfrac{1}{3}\cdot\left(\dfrac{5}{4}+\dfrac{7}{4}\right)-1=\dfrac{1}{3}\cdot\dfrac{12}{4}-1=\dfrac{1}{3}\cdot3-1=1-1=0\\ d.\dfrac{5}{13}+\left(-\dfrac{5}{17}\right)+\dfrac{-21}{41}+\dfrac{8}{13}+\dfrac{-20}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(-\dfrac{5}{17}\right)+\left(\dfrac{-21}{41}+\dfrac{-20}{41}\right)=1+\left(-\dfrac{5}{17}\right)-1=-\dfrac{5}{17}\)

\(e.\dfrac{27}{13}:\dfrac{9}{7}+\dfrac{12}{13}:\dfrac{9}{7}=\dfrac{27}{13}\cdot\dfrac{7}{9}+\dfrac{12}{13}\cdot\dfrac{7}{9}=\dfrac{7}{9}\cdot\left(\dfrac{27}{13}+\dfrac{12}{13}\right)=\dfrac{7}{9}\cdot\dfrac{39}{13}=\dfrac{7}{9}\cdot3=\dfrac{7}{3}\\ g.\dfrac{8}{15}\cdot-\dfrac{4}{9}+\dfrac{8}{15}:\dfrac{-9}{5}-3\dfrac{2}{5}=\dfrac{8}{15}\cdot\dfrac{-4}{9}+\dfrac{8}{15}\cdot\dfrac{-5}{9}-\dfrac{17}{5}=\dfrac{8}{15}\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)-\dfrac{17}{5}=-\dfrac{8}{15}-\dfrac{17}{5}=-\dfrac{59}{15}\\ h.\left(-\dfrac{2}{3}+\dfrac{3}{13}\right):\dfrac{7}{8}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right):\dfrac{7}{8}=\left(-\dfrac{2}{3}+\dfrac{3}{13}\right)\cdot\dfrac{8}{7}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right)\cdot\dfrac{8}{7}=\dfrac{8}{7}\cdot\left(-\dfrac{2}{3}+\dfrac{3}{13}-\dfrac{1}{3}+\dfrac{10}{3}\right)=\dfrac{8}{7}\cdot\left(-1+1\right)=\dfrac{8}{7}\cdot0=0\)

Ta có: ΔAHD vuông tại H

=>AD là cạnh huyền

=>AD>AH

mà AD=BC(ABCD là hình thang cân)

nên BC>AH

Ta có: KI là đường trung trực của AH

=>KI\(\perp\)AH và K là trung điểm của AH

Ta có: KI\(\perp\)AH

AH\(\perp\)HD

Do đó: KI//HD

=>\(\widehat{KIH}=\widehat{IHD}\)(1)

Xét ΔAHD có

K là trung điểm của AH

KI//HD

Do đó: I là trung điểm của AD

ΔAHD vuông tại H

mà HI là đường trung tuyến

nên IH=ID

=>ΔIHD cân tại I

=>\(\widehat{IHD}=\widehat{IDH}=\widehat{ADC}\left(2\right)\)

ABCD là hình thang cân

=>\(\widehat{ADC}=\widehat{BCD}\)(hai góc kề đáy CD)(3)

Từ (1),(2),(3) suy ra \(\widehat{HIK}=\widehat{BCD}\)

\(a.\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:-\dfrac{9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\cdot\dfrac{-2}{9}=\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

\(b.\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{4\cdot2-5}{10}=\dfrac{1}{2}:\dfrac{3}{10}=\dfrac{1}{2}\cdot\dfrac{10}{3}=\dfrac{5}{3}\)

\(c.\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}:\dfrac{5}{9}-1=\dfrac{1}{9}\cdot\dfrac{9}{5}-1=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

\(d.\left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}-6\dfrac{1}{2}\right)=\dfrac{9}{25}-\dfrac{4}{5}+6\dfrac{1}{2}=\dfrac{9}{25}-\dfrac{20}{25}+\dfrac{13}{2}=\dfrac{-11}{25}+\dfrac{13}{2}=\dfrac{303}{50}\)

a) \(\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:\dfrac{-9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\times\dfrac{2}{-9}\\ =\dfrac{1}{4}-\left(-\dfrac{1}{12}\right)=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

b) \(\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{3}{10}\\ =\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

c) \(\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}\times\dfrac{9}{5}-1=\dfrac{1}{5}-1\\ =-\dfrac{4}{5}\)

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

=>\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)

\(\left(2\dfrac{3}{4}-\dfrac{1}{2}\right):\dfrac{3}{5}-3\)

\(=\left(2+\dfrac{1}{4}\right):\dfrac{3}{5}-3=\dfrac{9}{4}\cdot\dfrac{5}{3}-3\)

\(=\dfrac{45}{12}-3=\dfrac{45}{12}-\dfrac{36}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)