\(17,a,\dfrac{3}{5}.-\dfrac{2}{9}-\dfrac{7}{9}.\dfrac{3}{5}=\dfrac{3}{5}.\left(\dfrac{-2}{9}-\dfrac{7}{9}\right)=\dfrac{3}{5}.-1=-\dfrac{3}{5}\)
\(b,\left(-\dfrac{1}{3}\right)^{2023}\div\left(-\dfrac{1}{3}\right)^{2021}=\left(-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(c,\sqrt{16}+\sqrt{9}-\sqrt{25}=4+3-5=7-5=2\)
`18`
\(a,x+0,8=-0,2\)
\(x=-0,2-0,8\)
\(x=-1\)
\(b,2x-\dfrac{1}{3}=\dfrac{5}{3}\)
\(2x=\dfrac{5}{3}+\dfrac{1}{3}\)
\(2x=2\)
\(x=2\div2=1\)
`19,`
`a,` vì `AC // DE =>` \(\widehat{ACE} =\widehat{DEB}=60^0\)
\(b,\) vì `AC // DE =>` \(\widehat{CAD} +\widehat{ADE}=180^0\) (2 góc trong cùng phía)
\(90^0+\widehat{ADE}=180^0\)
`=>` \(\widehat{ADE} =90^0\)
`=>` \(AB\perp DE\)
\(c,\)Vì `AC//DE =>` \(\widehat{CAD}=\widehat{EDB}=90^0\)
Xét tam giác DEB có:
\(\widehat{D}+\widehat{B}+\widehat{E}=180^0\)
\(90^0+\widehat{B}+60^0=180^0\)
`=>` \(\widehat{B}=30^0\)
`20,`
Xét tam giác ABD và tam giác ACD có:
`AB = AC (gt)`
AD chung
`DB = DC (gt)`
`=>` Tam giác ABD = Tam giác ACD (c-c-c)