Giúp mik vs mng
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2:
a: Xét tứ giác DIHK có
\(\widehat{DIH}=\widehat{DKH}=\widehat{IDK}=90^0\)
Do đó: DIHK là hình chữ nhật
Suy ra: DH=KI(1)
Xét ΔDEF vuông tại D có DH là đường cao ứng với cạnh huyền EF
nên \(DH^2=HE\cdot HF\left(2\right)\)
Từ (1) và (2) suy ra \(IK^2=HE\cdot HF\)
a, Theo tc 2 tt cắt nhau: \(AE=EC;BF=CF\)
Vậy \(AE+BF=EC+CF=EF\)
b, Vì \(\left\{{}\begin{matrix}AE=EC\\\widehat{EAO}=\widehat{ECO}=90^0\\OE.chung\end{matrix}\right.\) nên \(\Delta AOE=\Delta COE\)
\(\Rightarrow\widehat{AOE}=\widehat{EOC}\) hay OE là p/g \(\widehat{AOC}\)
Cmtt: \(\Delta BOF=\Delta COF\Rightarrow\widehat{BOF}=\widehat{COF}\) hay OF là p/g \(\widehat{BOC}\)
Vậy \(\widehat{EOF}=\widehat{COF}+\widehat{COE}=\dfrac{1}{2}\left(\widehat{AOC}+\widehat{BOC}\right)=90^0\) hay OE⊥OF
1 will come
2 drinks
3 go
4 visited
5 is playing
6 won't invite
7 will be
8 are waiting
9 drawing
10 playing
11 wasn't
12 made
13 is writing
14 takes
15 was
16 didn't meet
17 going
18 teaches
19 don't go
20 will visit
\(a^3+3a^2b+3ab^2+b^3-2022=\left(a+b\right)^3-2022=\left(2021-2020\right)^3-2022=1-2022=-2021\)
1: \(75^3:\left(-25\right)^3=\left(\dfrac{75}{-25}\right)^3=\left(-3\right)^3=-27\)
2: \(\left(-60\right)^2:\left(-5\right)^2=\dfrac{60^2}{5^2}=12^2=144\)
3: \(169^2:\left(-13\right)^2=\dfrac{169^2}{13^2}=\left(\dfrac{169}{13}\right)^2=13^2=169\)
4: \(\left(\dfrac{1}{2}\right)^2:\left(\dfrac{3}{2}\right)^2=\left(\dfrac{1}{2}:\dfrac{3}{2}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
5: \(\left(\dfrac{2}{3}\right)^3:\left(\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}:\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}\cdot\dfrac{27}{8}\right)^3=\left(\dfrac{9}{4}\right)^3=\dfrac{729}{64}\)
6: \(\left(\dfrac{5}{4}\right)^4:\left(\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}:\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}\cdot\dfrac{2}{15}\right)^4=\left(\dfrac{1}{6}\right)^4=\dfrac{1}{1296}\)
7: \(\left(\dfrac{7}{8}\right)^5:\left(\dfrac{21}{16}\right)^5\)
\(=\left(\dfrac{7}{8}:\dfrac{21}{16}\right)^5\)
\(=\left(\dfrac{7}{8}\cdot\dfrac{16}{21}\right)^5=\left(\dfrac{2}{3}\right)^5=\dfrac{32}{243}\)
8: \(\left(\dfrac{5}{6}\right)^4:\left(\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}:\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}\cdot\dfrac{18}{25}\right)^4=\left(\dfrac{3}{5}\right)^4=\dfrac{81}{625}\)
9:
\(\left(-\dfrac{3}{4}\right)^3:\left(\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}:\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}\cdot\dfrac{8}{9}\right)^3\)
\(=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
10:
\(\left(\dfrac{9}{10}\right)^6:\left(\dfrac{27}{-20}\right)^6=\left(\dfrac{9}{10}:\dfrac{-27}{20}\right)^6\)
\(=\left(\dfrac{9}{10}\cdot\dfrac{20}{-27}\right)^6=\left(-\dfrac{2}{3}\right)^6=\dfrac{64}{729}\)
e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x+2}{x+3}\)
a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)