(IMAS) The houses of Mary and Jerry are connected by a trail. One day, they started from their house at the same time and walked towards the other's house. The speed of Mary is 1.5 times that of Jerry and they met 12 minutes later. On the next day, Mary left his house and walked to Jerry's house at the same speed. How long would it take Mary to reach Jerry's house?
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\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)
\(-2x^2=-32\)
\(\Rightarrow x^2=\dfrac{-32}{-2}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy: ...
\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
\(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=\left(3-5-6\right)-\left(\dfrac{1}{4}-\dfrac{7}{4}\right)-\dfrac{3}{2}+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{6}{5}\)
\(=-8-\dfrac{6}{4}-\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{6}{5}\)
\(=-8-\dfrac{3}{2}-\dfrac{3}{2}+1+\dfrac{6}{5}\)
\(=-7-3+\dfrac{6}{5}\)
\(=\dfrac{6}{5}-10\)
\(=-\dfrac{44}{5}\)
Để A = 0 thì |x² - 1| = 0 và |x + 1| = 0
*) |x² - 1| = 0
x² - 1 = 0
x² = 1
x = 1 hoặc x = -1 (1)
*) |x + 1| = 0
x + 1 = 0
x = -1 (2)
Từ (1) và (2) ⇒ x = -1
Vậy x = -1 thì A = 0
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Qua C vẽ tia Cz // DE
⇒ ∠DCz = ∠CDE = 108⁰
⇒ ∠zCA = ∠DCz - ∠DCA
= 108⁰ - 48⁰
= 60⁰
⇒ ∠zCA = ∠CAB = 60⁰
Mà ∠zCA và ∠CAB là hai góc so le trong
⇒ Cz // AB
Lại có:
Cz // DE
⇒ AB // DE
\(\dfrac{-3}{4}< \dfrac{a}{12}< \dfrac{-5}{9}\)
\(\Rightarrow\dfrac{-27}{36}< \dfrac{3a}{36}< \dfrac{-20}{36}\)
\(\Rightarrow-27< 3a< -20\)
\(\Rightarrow a=\left\{-8;-7\right\}\)
\(\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{2^{-1}}}\)
\(=\dfrac{1+\dfrac{1}{2^{-1}}+1-\dfrac{1}{2^{-1}}}{\left(1-\dfrac{1}{2^{-1}}\right)\left(1+\dfrac{1}{2^{-1}}\right)}\)
\(=\dfrac{2}{\left(1-2\right)\left(1+2\right)}=\dfrac{2}{1-4}=-\dfrac{2}{3}\)