Tính một cách hợp lí:

a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111};\)

b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19};\)

c) \(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}.\)

Tính: 

a) \(\dfrac{5}{16}-\dfrac{5}{24};\)

b) \(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)\);

c) \(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right).\)

Tìm số đối của mỗi phân số sau: \(\dfrac{9}{25};\dfrac{-8}{27};-\dfrac{15}{31};\dfrac{-3}{-5};\dfrac{5}{-6}.\)

Tính một cách hợp lí:

a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\);

b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6};\)

c) \(\dfrac{-5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}.\)

Tính:

a) \(\dfrac{-2}{9}+\dfrac{7}{-9};\)        b) \(\dfrac{1}{-6}+\dfrac{13}{-15};\)        c) \(\dfrac{5}{-6}+\dfrac{-5}{12}+\dfrac{7}{18}.\)

a) \(\widehat{xOy}< \widehat{xOz}\), và hai tia Oy, Oz nằm trên cùng một nửa mặt phẳng bờ chứa tia Ox, do đó tia Oy nằm giữa hai tia Ox và Oz.

b) \(\widehat{zOy}=\widehat{xOz}-\widehat{xOy}=80^o-30^o=50^o.\)

c) Vì \(\widehat{xOy}\ne\widehat{yOz}\) nên tia Oy không là tia phân giác của góc xOz.

d) \(\widehat{x'Oz}=180^o-80^o=100^o\Rightarrow\widehat{mOz}=\dfrac{\widehat{x'Oz}}{2}=\dfrac{100^o}{2}=50^o.\)

\(\widehat{xOz}=80^o\\ \Rightarrow\widehat{mOz}< \widehat{xOz}.\)

b)

\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)

a)

\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100.}\)