Không có dữ kiện để tính toán tổng số sách 4 tổ đã góp. Bạn xem lại đề.

Có 3 góc

aOb;  cOb;   cOd

Nha bạn

\(A=\dfrac{2}{3}+\dfrac{4}{6}:\dfrac{8}{9}\)

\(=\dfrac{2}{3}+\dfrac{2}{3}\cdot\dfrac{9}{8}\)

\(=\dfrac{2}{3}\left(1+\dfrac{9}{8}\right)=\dfrac{2}{3}\cdot\dfrac{17}{8}=\dfrac{17}{12}\)

\(\dfrac{13\cdot\left(-135\right)+13\cdot115}{75-95}=\dfrac{113\left(-135+115\right)}{-20}\)

\(=\dfrac{113\cdot\left(-20\right)}{-20}=113\)

Đề bài yêu cầu gì vậy bạn?

= gg ;))

\(C=\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\)

=>\(4C=2+\dfrac{1}{2}+...+\dfrac{1}{2^{97}}\)

=>\(4C-C=2+\dfrac{1}{2}+...+\dfrac{1}{2^{97}}-\dfrac{1}{2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{99}}\)

=>\(3C=2-\dfrac{1}{2^{99}}=\dfrac{2^{100}-1}{2^{99}}\)

=>\(C=\dfrac{2^{100}-1}{3\cdot2^{99}}\)

cảm ơn bạn .Chúc bạn học giỏi !

Bài 10:

a: \(x=\dfrac{-1}{4}+\dfrac{2}{13}\)

=>\(x=\dfrac{-13}{52}+\dfrac{8}{52}=-\dfrac{5}{52}\)

b: \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)

=>\(\dfrac{x}{3}=\dfrac{14-3}{21}=\dfrac{11}{21}\)

=>\(x=\dfrac{11}{21}\cdot3=\dfrac{11}{7}\)

c: \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

=>\(-\dfrac{5}{6}-x=\dfrac{7-4}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

=>\(x=-\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{-10-3}{12}=-\dfrac{13}{12}\)

d: \(5\left(x-3\right)=\dfrac{5}{7}\)

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

=>\(x=3+\dfrac{1}{7}=\dfrac{22}{7}\)

e: \(-4\left(x-\dfrac{1}{3}\right)=\dfrac{-8}{3}\)

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

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

f: \(x+\dfrac{2}{3}=\dfrac{-1}{12}\cdot\dfrac{-4}{5}\)

=>\(x+\dfrac{2}{3}=\dfrac{1}{15}\)

=>\(x=\dfrac{1}{15}-\dfrac{2}{3}=\dfrac{1}{15}-\dfrac{10}{15}=-\dfrac{9}{15}=-\dfrac{3}{5}\)

g: \(x-4=\dfrac{-14}{35}:\dfrac{7}{5}\)

=>\(x-4=-\dfrac{2}{5}\cdot\dfrac{5}{7}=-\dfrac{2}{7}\)

=>\(x=4-\dfrac{2}{7}=\dfrac{26}{7}\)

h: \(-\dfrac{3}{7}x=\dfrac{3}{56}\cdot\dfrac{28}{9}\)

=>\(-\dfrac{3}{7}\cdot x=\dfrac{3}{9}\cdot\dfrac{28}{56}=\dfrac{1}{6}\)

=>\(x=-\dfrac{1}{6}:\dfrac{3}{7}=\dfrac{-7}{18}\)

i: \(\dfrac{1}{4}x-\dfrac{7}{5}=\dfrac{-5}{8}:\dfrac{15}{4}\)

=>\(\dfrac{1}{4}x-\dfrac{7}{5}=\dfrac{-5}{8}\cdot\dfrac{4}{15}=\dfrac{-1}{6}\)

=>\(\dfrac{x}{4}=\dfrac{-1}{6}+\dfrac{7}{5}=\dfrac{-5+42}{30}=\dfrac{37}{30}\)

=>\(x=\dfrac{37}{30}\cdot4=37\cdot\dfrac{2}{15}=\dfrac{74}{15}\)

k: \(x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\)

=>\(x-\dfrac{3}{10}=\dfrac{7}{15}\cdot\dfrac{5}{3}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}+\dfrac{3}{10}=\dfrac{70+27}{90}=\dfrac{97}{90}\)

l: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{11}{9}\)

=>\(x+\dfrac{3}{22}=\dfrac{27}{9}\cdot\dfrac{11}{121}=\dfrac{3}{11}\)

=>\(x=\dfrac{3}{11}-\dfrac{3}{22}=\dfrac{3}{22}\)

m: \(\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)

=>\(\dfrac{1}{5}:x=\dfrac{2}{35}\)

=>\(x=\dfrac{1}{5}:\dfrac{2}{35}=\dfrac{1}{5}\cdot\dfrac{35}{2}=\dfrac{7}{2}\)

\(A=4+2^2+2^3+...+2^{20}\)

Đặt \(B=2^2+2^3+...+2^{20}\)

=>\(2B=2^3+2^4+...+2^{21}\)

=>\(2B-B=2^3+2^4+...+2^{21}-2^2-...-2^{20}\)

=>\(B=2^{21}-2^2\)

=>\(A=4+2^{21}-2^2=2^{21}\)