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\(\dfrac{1}{430}+\dfrac{1}{324}\)

\(=\dfrac{162}{69660}+\dfrac{215}{69660}\)

\(=\dfrac{377}{69660}\)

\(\dfrac{1}{555}+\dfrac{1}{678}\)

\(=\dfrac{678}{376290}+\dfrac{555}{376290}\)

\(=\dfrac{1233}{376290}=\dfrac{411}{125430}=\dfrac{137}{41810}\)

\(360:24=15\)

Đặt tính:

360         24

12           15

120

    0

\(\)\(a.\) Xét \(\text{△ }ABC;\text{△ }HBA\) có:

\(\widehat{B}\) \(chung;\widehat{BHA}=\widehat{BAC}=90^0\)

\(\Rightarrow\text{△ }ABC\text{ }\text{∼ △}HBA\left(g.g\right)\)

\(b.\) Áp dụng định lý Pytago:

 \(BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25\left(cm\right)\)

\(-AH=2S_{ABC}:BC=AB\cdot AC:BC=15\cdot20:25=12\left(cm\right)\) (không dùng định lý)

\(-BH=\sqrt{AB^2-AH^2}=\sqrt{15^2-12^2}=9\left(cm\right)\) (Áp dụng định lý Pytago).

\(c.\) Áp dụng tính chất đường phân giác:

\(\dfrac{DA}{DC}=\dfrac{AB}{BC}=\dfrac{15}{25}=\dfrac{3}{5}\)

\(DA+DC=AC=20\)

\(\Rightarrow DA=20:\left(3+5\right)\cdot3=7,5\left(cm\right)\)

\(DC=AC-DA=20-7,5=12,5\left(cm\right)\)

\(a.\) \(\dfrac{42}{54}=\dfrac{7}{x}\)

\(\Rightarrow x\cdot42=7\cdot54\)

\(\Rightarrow x\cdot42=378\)

\(\Rightarrow x=378:42\)

\(\Rightarrow x=9\)

Vậy \(\dfrac{42}{54}=\dfrac{7}{9}.\)

\(b.\) \(\dfrac{-2}{3}=\dfrac{y}{15}\)

\(\Rightarrow y\cdot3=\left(-2\right)\cdot15\)

\(\Rightarrow y\cdot3=\left(-30\right)\)

\(\Rightarrow y=\left(-30\right):3\)

\(\Rightarrow y=\left(-10\right)\)

Vậy \(\dfrac{-2}{3}=\dfrac{-10}{15}\)

\(c.\) \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)

\(\Rightarrow x\cdot6=3\cdot10\)

\(\Rightarrow x\cdot6=30\)

\(\Rightarrow x=30:6\)

\(\Rightarrow x=5\)

Vậy: \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{y}{-20}\)

Mặt khác: \(\dfrac{3}{5}=\dfrac{y}{-20}\)

\(\Rightarrow y\cdot5=3\cdot\left(-20\right)\)

\(\Rightarrow y\cdot5=\left(-60\right)\)

\(\Rightarrow y=\left(-60\right):5\)

\(\Rightarrow y=\left(-12\right)\)

Vậy \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{-12}{-20}\)

\(d.\) \(\dfrac{-x}{-6}=\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\)

Do cùng mẫu số nên ta xét tử, ta thấy:

\(x=\left(-5\right)\)

Vậy \(\dfrac{-5}{6}=\dfrac{-5}{6}\)

Ta có: Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{78}\)

\(\dfrac{A}{2}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{20}+...+\dfrac{1}{156}\)

\(\dfrac{A}{2}=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{12\times13}\)

\(\dfrac{A}{2}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{12}-\dfrac{1}{13}\)

\(\dfrac{A}{2}=\dfrac{1}{1}-\dfrac{1}{13}\)

\(\dfrac{A}{2}=\dfrac{13}{13}-\dfrac{1}{13}\)

\(\dfrac{A}{2}=\dfrac{12}{13}\)

\(A=\dfrac{12}{13}\times2\)

\(A=\dfrac{24}{13}\)

Vậy \(A=\dfrac{24}{13}\)

Ta có: \(8b-9a=31\)

\(\Rightarrow8b=31+9a\)

\(\Rightarrow b=\dfrac{31+9a}{8}\)

\(\Rightarrow b=\dfrac{32+8a+a-1}{8}\)

\(\Rightarrow b=\dfrac{8\cdot\left(4+a\right)+a-1}{8}\)

\(\Rightarrow b=4+a+\dfrac{a-1}{8}\)

Để \(b\in N\) thì:

\(\dfrac{a-1}{8}\in N\)

\(\Rightarrow a-1⋮8\)

\(\Rightarrow a-1=8k\left(k\in N\right)\)

\(\Rightarrow a=8k+1\)

Khi đó: \(b=4+8k+1+\dfrac{8k+1-1}{8}\)

\(\Rightarrow b=5+8k+\dfrac{8k}{8}\)

\(\Rightarrow b=5+8k+k\)

\(\Rightarrow b=5+9k\)

Mặt khác: \(\dfrac{11}{17}< \dfrac{a}{b}< \dfrac{23}{29}\)

\(\Rightarrow\dfrac{11}{17}< \dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)

Xét: \(\dfrac{11}{17}< \dfrac{8k+1}{5+9k}\)

\(\Rightarrow11\left(5+9k\right)< 17\left(8k+1\right)\)

\(\Rightarrow55+99k< 136k+17\)

\(\Rightarrow136k-99k>55-17\)

\(\Rightarrow37k>38\)

\(\Rightarrow k>\dfrac{38}{37}\left(1\right)\)

Xét: \(\dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)

\(\Rightarrow29\left(8k+1\right)< 23\left(5+9k\right)\)

\(\Rightarrow232k+29< 115+207k\)

\(\Rightarrow232k-207k< 115-29\)

\(\Rightarrow25k< 86\)

\(\Rightarrow k< \dfrac{86}{25}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{38}{27}< k< \dfrac{86}{25}\)

Mà \(k\in N\)

\(\Rightarrow k\in\left\{2;3\right\}\)

\(+,\) \(k=2\).

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot2+1=17\\b=5+9\cdot2=23\end{matrix}\right.\)

\(+,\) \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot3+1=25\\b=5+9\cdot3=32\end{matrix}\right.\)

\(\Rightarrow\) Vậy \(\left(a;b\right)\in\left\{\left(17;23\right),\left(25;32\right)\right\}\)