a) \(\dfrac{7}{12}< \dfrac{7+1}{12+1}< \dfrac{78}{13}\Rightarrow\dfrac{7}{12}< \dfrac{8}{13}\)

b) \(-4,25=-\dfrac{425}{100}=-\dfrac{17}{4}=-\dfrac{34}{8}< -\dfrac{28}{8}\Rightarrow-4,25< -\dfrac{28}{8}\)

c) \(-0,33>-0,5=-\dfrac{1}{2}=-\dfrac{19}{38}\Rightarrow-0,33>-\dfrac{19}{38}\)

d) \(\dfrac{11}{13}< \dfrac{11+2}{13+2}=\dfrac{13}{15}\Rightarrow\dfrac{11}{13}< \dfrac{13}{15}\Rightarrow-\dfrac{11}{13}>-\dfrac{13}{15}\)

a) Xét tam giác ACB và ADC, có \(\widehat{A}\) chung và \(\widehat{ACB}=\widehat{ADC}\left(gt\right)\), suy ra đpcm.

b) Từ câu a) \(\Rightarrow\dfrac{AB}{AC}=\dfrac{AC}{AD}\) \(\Rightarrow AC^2=AB.AD\)

Kẻ phân giác BE của tam giác ABC. Vì \(\widehat{B}=2\widehat{C}\)  nên \(\widehat{ABE}=\widehat{ADC}\) hay BE//CD. Mặt khác, \(\dfrac{EA}{EC}=\dfrac{BA}{BC}=\dfrac{4}{5}\) nên suy ra \(\dfrac{BA}{BD}=\dfrac{4}{5}\Leftrightarrow\dfrac{4}{BD}=\dfrac{4}{5}\Leftrightarrow BD=5\),  suy ra \(AD=AB+BD=4+5=9\).

\(\Rightarrow AC^2=AB.AD=4.9=36\) \(\Rightarrow AC=6\).

Vậy \(AC=6\)

 Dạ thưa cô, cái này em áp dụng tính chất đường phân giác trong tam giác ạ. Cái này lớp 9 được dùng luôn không cần chứng minh ạ.

Bài 1: a, \(\dfrac{4}{35}\) + \(\dfrac{8}{21}\) = \(\dfrac{12}{105}\) + \(\dfrac{40}{105}\) = \(\dfrac{52}{105}\)

          b, \(\dfrac{21}{49}\) + \(\dfrac{9}{27}\) = \(\dfrac{567}{1323}\) + \(\dfrac{441}{1323}\) = \(\dfrac{1008}{1323}\) = \(\dfrac{16}{21}\)

          c, \(\dfrac{26}{15}\) - \(\dfrac{7}{5}\) = \(\dfrac{26}{15}\) - \(\dfrac{21}{15}\) = \(\dfrac{5}{15}\) = \(\dfrac{1}{3}\)

          d, \(\dfrac{29}{36}\) - \(\dfrac{2}{9}\) = \(\dfrac{29}{36}\) - \(\dfrac{8}{36}\) = \(\dfrac{21}{36}\) = \(\dfrac{7}{12}\)

          e, \(\dfrac{3}{7}\)\(\times\)\(\dfrac{14}{15}\) = \(\dfrac{2}{5}\)

         f, \(\dfrac{35}{9}\): \(\dfrac{7}{81}\) = \(\dfrac{35}{9}\) \(\times\) \(\dfrac{81}{7}\) = 45

         g,\(\dfrac{28}{17}\) \(\times\) \(\dfrac{68}{14}\) = 8

        h, \(\dfrac{35}{46}\): \(\dfrac{205}{23}\) = \(\dfrac{35}{46}\) \(\times\) \(\dfrac{23}{205}\) = \(\dfrac{7}{82}\)

P = 2\(x\)³ + 3\(x\)² + 4\(x\) + 5 

Thay \(x\) = 3 vào P ta có: 

P = 2.3³ + 3.3² + 4.3 + 5

P = 54 + 27 + 12 + 5

P = 98

dễ mà

\(S=1+3^2+3^3+3^4+...+3^{2023}\left(1\right)\)

Ta có \(S+3=1+3+3^2+3^3+3^4+...+3^{2023}\)

\(\Rightarrow S+3=\dfrac{3^{2023+1}-1}{3-1}=\dfrac{3^{2024}-1}{2}\)

\(\Rightarrow S=\dfrac{3^{2024}-1}{2}-3==\dfrac{3^{2024}-7}{2}\)

\(S=1+3^2+3^3+3^4+...+3^{2023}\\ 3S=3+3^3+3^4+3^5+...+3^{2024}\\ 3S-S=3+3^3+3^4+3^5+...+3^{2024}-1-3^2-3^3-3^4-...-3^{2023}\\ 2S=3+3^{2024}-1-3^2\\ 2S=3+3^{2024}-1-9\\ 2S=-3+3^{2024}\\ S=\dfrac{-3+3^{2024}}{2}\)

\(x-8:2=-4\\ x-4=-4\\ x=-4+4\\ x=0\\ -5.x-7=-17\\ -5.x=-17+7\\ -5.x=-10\\ x=-10:\left(-5\right)\\ x=2\\ 7.\left(4-x\right)< 0\\ 4-x< 0:7\\ 4-x< 0\\ x\in\left\{5;6;7;8;...\right\}\)

a) \(...\Rightarrow x-4=-4\Rightarrow x=-4+4-0\)

b) \(...\Rightarrow-5x=-17+7\Rightarrow-5x=-10\Rightarrow x=\left(-10\right):\left(-5\right)=2\)

c) \(...\Rightarrow4-x< 0\Rightarrow x>4\)

A = \(\dfrac{254\times399-145}{254+399\times253}\)

A = \(\dfrac{\left(253+1\right)\times399-145}{254+399\times253}\)

A = \(\dfrac{253\times399+399-145}{254+399\times253}\)

A = \(\dfrac{253\times399+254}{254+399\times253}\)

A = 1 

a) \(254x399-145=\left(250+4\right)\left(400-1\right)-145\)

\(=100000+1600-250-145-4=101600-\left(250+145+4\right)\)

\(=101600-399=101201\)

b) \(254+399x253=254+\left(400-1\right)x\left(250+3\right)\)

\(=254+100000+1200-250-3\)

\(=101454-250-3=101201\)