từ điểm M nằm ngoài đường tròn (o) vẽ 2 tiếp tuyến MA MB với đường tròn (o).Lấy N bất kì thuộc cung nhỏ ab.Tiếp tuyến tại N của (O) cắt MA,MB tại P,Q. a) chứng minh tứ giác APNO nội tiếp. b) Chứng minh AOB=2POQ. c)Gọi H là hình chiếu vuông góc của N lên AB. CM HN là tia phân giác của PHQ
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Bài 1:
a: \(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)
\(=\left(-\dfrac{3}{5}-\dfrac{4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\)
\(=\dfrac{7}{21}=\dfrac{1}{3}\)
b: \(-\dfrac{3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
\(=-\dfrac{3}{17}+\dfrac{3}{17}+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}=\dfrac{2}{3}\)
c: \(\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(-\dfrac{5}{21}-\dfrac{16}{21}\right)+1\)
=-1+1=0
d: \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
\(=\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
\(=-1+1=0\)
e: \(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
f: \(\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}\)
\(=-\dfrac{3}{8}-\dfrac{5}{8}=-1\)
g: \(\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}=\dfrac{-4}{-4}\cdot\dfrac{11}{11}\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
h: \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}\)
\(=\dfrac{7}{36}+\dfrac{32}{36}-\dfrac{24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
i: \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{32}{56}-\dfrac{35}{56}-\dfrac{6}{56}\)
\(=-\dfrac{9}{56}\)
l: \(\dfrac{-6}{11}:\left(\dfrac{3}{5}\cdot\dfrac{4}{11}\right)\)
\(=-\dfrac{6}{11}:\dfrac{12}{55}\)
\(=-\dfrac{6}{11}\cdot\dfrac{55}{12}=\dfrac{-5}{2}\)
Dạng 2:
a: \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
=>\(x\cdot\dfrac{7}{4}-5=\dfrac{10}{3}\)
=>\(x\cdot\dfrac{7}{4}=\dfrac{10}{3}+5=\dfrac{25}{3}\)
=>\(x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{25}{3}\cdot\dfrac{4}{7}=\dfrac{100}{21}\)
b: \(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\)
=>\(\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{7}{12}-\dfrac{3}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
=>\(x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
c: \(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\)
=>\(\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
=>\(x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
=>\(x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
d: \(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\)
=>\(\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
=>\(3x=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{1}{3}\cdot\dfrac{4}{-21}=\dfrac{-4}{63}\)
=>\(x=-\dfrac{4}{63}:3=-\dfrac{4}{189}\)
e: \(2x^2-72=0\)
=>\(2x^2=72\)
=>\(x^2=36\)
=>\(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
f: \(\left(\dfrac{3}{5}x-0,75\right):\dfrac{3}{7}=2\dfrac{4}{5}\)
=>\(\left(\dfrac{3}{5}x-0,75\right):\dfrac{3}{7}=\dfrac{14}{5}\)
=>\(\dfrac{3}{5}x-0,75=\dfrac{14}{5}\cdot\dfrac{3}{7}=\dfrac{6}{5}\)
=>\(\dfrac{3}{5}x=\dfrac{6}{5}+\dfrac{3}{4}=\dfrac{39}{20}\)
=>\(x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{39}{20}\cdot\dfrac{5}{3}=\dfrac{13}{4}\)
g: \(2x+\dfrac{3}{10}=1\dfrac{5}{6}\cdot\dfrac{6}{11}\)
=>\(2x+\dfrac{3}{10}=\dfrac{11}{6}\cdot\dfrac{6}{11}=1\)
=>\(2x=\dfrac{7}{10}\)
=>\(x=\dfrac{7}{20}\)
h: \(2\dfrac{1}{4}:\left(x-7\dfrac{1}{3}\right)=-1,5\)
=>\(\dfrac{9}{4}:\left(x-\dfrac{22}{3}\right)=-\dfrac{3}{2}\)
=>\(x-\dfrac{22}{3}=\dfrac{-9}{4}:\dfrac{3}{2}=-\dfrac{9}{4}\cdot\dfrac{2}{3}=\dfrac{-3}{2}\)
=>\(x=-\dfrac{3}{2}+\dfrac{22}{3}=\dfrac{35}{6}\)
a: \(2x-\dfrac{5}{4}=\dfrac{3}{2}\)
=>\(2x=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{6}{4}+\dfrac{5}{4}=\dfrac{11}{4}\)
=>\(x=\dfrac{11}{8}\)
b: \(\dfrac{1}{5}:x-\dfrac{6}{7}=\dfrac{3}{14}\)
=>\(\dfrac{1}{5}:x=\dfrac{3}{14}+\dfrac{6}{7}=\dfrac{3}{14}+\dfrac{12}{14}=\dfrac{15}{14}\)
=>\(x=\dfrac{1}{5}:\dfrac{15}{14}=\dfrac{1}{5}\cdot\dfrac{14}{15}=\dfrac{14}{75}\)
c: \(x:\dfrac{4}{9}+\dfrac{5}{9}=\dfrac{13}{3}\)
=>\(x:\dfrac{4}{9}=\dfrac{13}{3}-\dfrac{5}{9}=\dfrac{39}{9}-\dfrac{5}{9}=\dfrac{34}{9}\)
=>\(x=\dfrac{34}{9}\cdot\dfrac{4}{9}=\dfrac{136}{81}\)
d: \(17-x\times\dfrac{8}{3}=\dfrac{1}{2}\)
=>\(x\times\dfrac{8}{3}=17-\dfrac{1}{2}=\dfrac{33}{2}\)
=>\(x=\dfrac{33}{2}:\dfrac{8}{3}=\dfrac{33}{2}\times\dfrac{3}{8}=\dfrac{99}{16}\)
e: \(\dfrac{21}{4}+x:\dfrac{5}{2}=\dfrac{3}{2}\)
=>\(x:\dfrac{5}{2}=\dfrac{3}{2}-\dfrac{21}{4}=\dfrac{-15}{4}\)
=>\(x=-\dfrac{15}{4}\times\dfrac{5}{2}=-\dfrac{75}{8}\)
g: \(\dfrac{18}{2}:2-4:x=\dfrac{3}{10}\)
=>\(4:x=\dfrac{9}{2}-\dfrac{3}{10}=\dfrac{45}{10}-\dfrac{3}{10}=\dfrac{42}{10}=\dfrac{21}{5}\)
=>\(x=4:\dfrac{21}{5}=\dfrac{20}{21}\)
a; 2\(x\) - \(\dfrac{5}{4}\) = \(\dfrac{3}{2}\)
2\(x\) = \(\dfrac{3}{2}\) + \(\dfrac{5}{4}\)
2\(x\) = \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{11}{4}\) : 2
\(x\) = \(\dfrac{11}{8}\)
Vì AD//BC
nên \(\dfrac{OA}{OC}=\dfrac{OD}{OB}=k\)
=>\(OC=k\times OA;OB=k\times OD\)
Vì \(OC=k\times OA\)
nên \(S_{DOC}=k\times S_{AOD}\)
Vì \(OB=k\times OD\)
nên \(S_{AOB}=k\times S_{AOD}\)
Do đó: \(S_{AOB}=S_{DOC}\)
Vì AD/BC
Nên OA/OC=OD/OB=k
=>OC=kxOA;OB=kxOD
Vì OC=kxOA
Nên sDOC=kx sAOD
Vì OB=kxOD
Nên sAOB=kx sAOD
Do đó:sAOB=sDOC
Tick cho mình nhé
\(\left\{{}\begin{matrix}8x-y=6\\x^2-y=-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x-y-x^2+y=6+6\\8x-y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-8x=-12\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-8x+12=0\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-2\right)\left(x-6\right)=0\\y=8x-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{2;6\right\}\\y=8x-6\end{matrix}\right.\)
Khi x=2 thì \(y=8\cdot2-6=16-6=10\)
Khi x=6 thì \(y=8\cdot6-6=42\)