\(3\dfrac{5}{9}=\dfrac{3\times9+5}{9}=\dfrac{32}{9}\\ 4\dfrac{7}{12}=\dfrac{4\times12+7}{12}=\dfrac{55}{12}\\ 5\dfrac{3}{10}=\dfrac{5\times10+3}{10}=\dfrac{53}{10}\\ 8\dfrac{13}{40}=\dfrac{8\times40+13}{40}=\dfrac{323}{40}\)

\(Bài13\\ a,x+\dfrac{4}{5}=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}-\dfrac{4}{5}\\ \Rightarrow x=-\dfrac{2}{15}\\ ----------\\ b,x+\dfrac{3}{2}=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{4}-\dfrac{3}{2}\\ \Rightarrow x=\dfrac{11-3.2}{4}\\ \Rightarrow x=\dfrac{5}{4}\\ ----------\\ c,\dfrac{25}{3}-x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{25}{3}-\dfrac{5}{6}\\ \Rightarrow x=\dfrac{15}{2}\)

\(Bài14\\ a,\dfrac{12}{7}+\dfrac{19}{7}+\dfrac{18}{7}=\dfrac{12+19+18}{7}=\dfrac{49}{7}=7\\ ----------\\ b,\dfrac{2}{5}+\dfrac{7}{12}+\dfrac{13}{12}=\dfrac{2}{5}+\dfrac{5}{3}=\dfrac{2.3+5.5}{15}=\dfrac{7}{3}\)

Sau bạn đăng từ 1 --> 2 câu th nhé 

a, Xét tứ giác \(BFED\) có

\(ED//FB\\ EF//BD\)

=> \(BFED\) là hình bình hành

b, Xét \(ABC\) có 

\(E\) là trung điểm \(AC\)

\(EF//BC\)

=> \(F\) là trung điểm \(AB\)

Xét tứ giác \(AHBD\) có

\(F\) là trung điểm \(AB\)

\(F\) là trung điểm \(HD\)

=> \(AHBD\) là hình bình hành 

=> \(HB//AD\) (đpcm)

\(x=7-4\sqrt{3}\\ =>\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\\ \)

Thay \(\sqrt{x}=2-\sqrt{3}\)

\(A=\dfrac{2-\sqrt{3}+2}{2-\sqrt{3}+3}=\dfrac{4-\sqrt{3}}{5-\sqrt{3}}=\dfrac{\left(4-\sqrt{3}\right)\left(5+\sqrt{3}\right)}{25-3}=\dfrac{20-5\sqrt{3}+4\sqrt{3}-3}{22}=\dfrac{17-\sqrt{3}}{22}\)

\(a,178=2.89\\ b,140=2^2.5.7\\ c,1075=5^2.43\\ d,2400=2^5.3.5^2\\ e,=10^5\)

\(1011-x=\dfrac{1}{2023}:\dfrac{2}{2023}\\ \Rightarrow1011-x=\dfrac{1}{2023}\times\dfrac{2023}{2}\\ \Rightarrow1011-x=\dfrac{1}{2}\\ \Rightarrow x=1011-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{2021}{2}\)

\(=\left(3-x\right)\left(3^2+3x+x^2\right)\\ =\left(3-x\right)\left(9+3x+x^2\right)\)

Áp dụng định lý tổng 4 góc trong tứ giác 

\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\\ \Rightarrow30^o+130^o+3\widehat{D}+\widehat{D}=360^o\\ \Rightarrow4\widehat{D}=360^o-130^o-30^o\\ 4\widehat{D}=200\\ \Rightarrow\widehat{D}=50^o\)

Do \(\widehat{C}=3\widehat{D}\\ \Rightarrow\widehat{C}=3\times50=150^o\)

\(15\\ a,-Đ\\ b,-S\\ c,-S\\d,-Đ \)

\(16,\)

Chiều rộng HCN là :

\(\dfrac{5}{3}-\dfrac{1}{2}=\dfrac{10-3}{6}=\dfrac{7}{6}\left(dm\right)\)

Diện tích HCN là :

\(S=\dfrac{5}{3}\times\dfrac{7}{6}=\dfrac{35}{18}\left(dm^2\right)\)

\(1,=\left|3-\sqrt{3}\right|=3-\sqrt{3}\\ ---------\\ 2,=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}=6\)