Kẻ Ct//Ax

\(\Rightarrow\widehat{BCt}=180-120=60^o\)(góc trong cùng phía)

\(\Rightarrow\widehat{tBA}=90-60=30^o\)

\(\widehat{CAx}=\widehat{tCA}=30^o\)(góc so le trong)

\(2.\left(x-3\right)^3=164\)

\(\left(x-3\right)^3=82=\left(\sqrt[3]{82}\right)^3\)

\(x-3=\sqrt[3]{82}\)

\(x=\sqrt[3]{82}+3\)

a) Ta có:

\(sin40=\dfrac{AB}{BC}=\dfrac{21}{BC}\)\(\Rightarrow BC=\dfrac{21}{sin40}\simeq33cm\)

\(cos40=\dfrac{AC}{BC}\Rightarrow AC=cos40.33\simeq25cm\)

b) \(sinB=\dfrac{AC}{BC}=\dfrac{25}{33}\Rightarrow\widehat{B}\simeq49^o\)

\(BD=\dfrac{2.BC.AB.cos24,5}{BC+AB}\simeq12cm\)

\(\left\{{}\begin{matrix}EF//AC\\BD\perp AC\end{matrix}\right.\)\(\Rightarrow EF\perp BD\)\(\Rightarrow\widehat{IMF}=90^o\)

Tương tự: \(\Rightarrow\widehat{INF}=90^o\)

=> \(IMFN\) là h.c.n

CM: tương tự

=> EFGH là hình chữ nhật

=> 4 điểm E, F, G, H cùng thuộc 1 đường tròn

\(\)

\(2^4.5-\left[131-\left(13-4\right)^2\right]+2020^0\)

\(=16.9-\left[131-9^2\right]+1\)

\(=90-\left(131-81\right)+1\)

\(=91-50=41\)

\(Ư\left(18\right)=\left(\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right)\)

\(\dfrac{3}{31}+\left(\dfrac{1}{6}+\dfrac{-5}{9}\right):\left(\dfrac{7}{24}-\dfrac{13}{18}\right)\)

\(=\dfrac{3}{31}+\dfrac{3-5.2}{18}:\dfrac{7.3-13.4}{72}\)

\(=\dfrac{3}{31}+\dfrac{-7}{18}:\dfrac{-31}{72}\)

\(=\dfrac{3}{31}+\dfrac{7.72}{18.31}=\dfrac{3}{31}+\dfrac{28}{31}=1\)

\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)\(\left(Đk:x,y\ne-1\right)\)

\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{2x}{x+1}+\dfrac{6y}{y+1}=-2\end{matrix}\right.\)

\(\Rightarrow\dfrac{5y}{y+1}=-5\)

\(\Leftrightarrow5y=-5y-5\)

\(\Leftrightarrow10y=-5\)

\(\Leftrightarrow y=-\dfrac{1}{2}\Rightarrow x=-2\)

Áp dụng định lý Pytago vào tam giác ABC

\(BC=\sqrt{3^2+4^2}=5\)

ÁP dụng hệ thức lượng vào tam giác ABC

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow AH=\sqrt{\dfrac{1}{\dfrac{1}{3^2}+\dfrac{1}{4^2}}}=\dfrac{12}{5}\)

\(sinB=\dfrac{AC}{BC}=\dfrac{4}{5}\)\(\Rightarrow\widehat{B}\simeq53,1^o\)