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Trong tam giác vuông ABP:
\(tanP=\dfrac{AB}{AP}\Rightarrow AP=\dfrac{AB}{tanP}\Rightarrow PQ+AQ=\dfrac{AB}{tanP}\) (1)
Trong tam giác vuông ABQ:
\(tanQ=\dfrac{AB}{AQ}\Rightarrow AQ=\dfrac{AB}{tanQ}\) (2)
\(\left(1\right);\left(2\right)\Rightarrow PQ+\dfrac{AB}{tanQ}=\dfrac{AB}{tanP}\Rightarrow PQ=AB\left(\dfrac{1}{tanP}-\dfrac{1}{tanQ}\right)\)
\(\Rightarrow AB=\dfrac{PQ}{\dfrac{1}{tanP}-\dfrac{1}{tanQ}}=\dfrac{100}{\dfrac{1}{tan15^0}-\dfrac{1}{tan55^0}}\approx33\left(m\right)\)
1 - x + 1/3 = x - 1/2
<=> 6(1-x) +2 = 6x - 3
<=> 6- 6x +2 = 6x -3
<=> 12x = 11
<=> x = 11/12
1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)
\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)
\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)
a: \(=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
b: \(=\dfrac{6+6\cdot4+6\cdot49}{15+15\cdot4+15\cdot49}=\dfrac{6}{15}=\dfrac{2}{5}\)
c: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)
\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9\)
\(a=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)\)
\(a=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)\)
\(a=2.7+2^4.7+2^7.7=7\left(2+2^4+2^7\right)⋮7\left(đpcm\right)\)
A=(2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)
A=2(1+2+2^2)+2^4(1+2+2^2)+2^7(1+2+2^2)
A=2.7+2^4.7+2^7.7\(⋮\)7
Vậy A\(⋮\)7
Đường tròn (C) tâm \(I\left(-2;-2\right)\) bán kính \(R=5\)
Gọi đường thẳng d qua A có dạng: \(a\left(x-6\right)+b\left(y-17\right)=0\)
\(\Leftrightarrow ax+by-6a-17b=0\) (\(a^2+b^2\ne0\))
d là tiếp tuyến của (C) khi và chỉ khi \(d\left(I;d\right)=R\)
\(\Leftrightarrow\dfrac{\left|-2a-2b-6a-17b\right|}{\sqrt{a^2+b^2}}=5\)
\(\Leftrightarrow\left|8a+19b\right|=5\sqrt{a^2+b^2}\)
\(\Leftrightarrow\left(8a+9b\right)^2=25\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(3a+4b\right)\left(13a+84b\right)=0\)
Chọn \(\left(a;b\right)=\left(4;-3\right);\left(84;-13\right)\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}4\left(x-6\right)-3\left(y-17\right)=0\\84\left(x-6\right)-13\left(y-17\right)=0\end{matrix}\right.\) \(\Leftrightarrow...\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-7}{156}\)
\(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-17}{12}\)
\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-7}{55}\)
\(\dfrac{-34}{37}.\dfrac{74}{-85}=\dfrac{4}{5}\)
\(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{10}{7}\)
Chúc bạn học tốt!!!
a) \(\left(-\dfrac{1}{39}\right)+\left(-\dfrac{1}{52}\right)=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
b) \(\left(-\dfrac{6}{9}\right)+\left(-\dfrac{12}{16}\right)=-\dfrac{6}{9}-\dfrac{12}{16}=-\dfrac{17}{12}\)
c) \(-\dfrac{2}{5}-\left(-\dfrac{3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}=-\dfrac{7}{55}\)
d) \(\left(-\dfrac{34}{37}\right)\cdot\left(-\dfrac{74}{85}\right)=2\cdot\dfrac{2}{5}=\dfrac{4}{5}\)
e) \(\left(-\dfrac{5}{9}\right):\left(-\dfrac{7}{18}\right)=\dfrac{5}{9}\cdot\dfrac{18}{7}=5\cdot\dfrac{2}{7}=\dfrac{10}{7}\)