Câu 58: B

Câu 59: C

40.B         41.A        42.A       43.D          44.B        45.A        46.D

Lời giải:

\(H=2-\frac{3}{17}-\frac{5}{23}+\frac{2}{17}-\frac{1}{2023}-\frac{16}{17}-\frac{18}{23}\\ =2-(\frac{3}{17}-\frac{2}{17}+\frac{16}{17})-(\frac{5}{23}+\frac{18}{23})-\frac{1}{2023}\\ =2-1-1-\frac{1}{2023}=-\frac{1}{2023}\)

---------------------

\(K=\frac{7}{23}.\frac{-11}{17}+\frac{7}{23}.\frac{4}{17}-\frac{7}{23}.\frac{10}{17}\\ =\frac{7}{23}(\frac{-11}{17}+\frac{4}{17}-\frac{10}{17})\\ =\frac{7}{23}.\frac{-17}{17}=\frac{-7}{23}\)

Xet tam giac BDC va tam giac CEB ta co 

^BDC = ^CEB = 90⁰

BC _ chung 

^BCD = ^CBE ( gt ) 

=> tam giac BDC = tam giac CEB ( ch - gn ) 

=> ^DBC = ^ECB ( 2 goc tuong ung ) 

Ta co ^B - ^DBC = ^ABD 

^C - ^ECB = ^ACE 

=> ^ABD = ^ACE 

Xet tam giac IBE va tam giac ICD 

^ABD = ^ACE ( cmt )

^BIE = ^CID ( doi dinh ) 

^BEI = ^IDC = 90⁰

Vay tam giac IBE = tam giac ICD (g.g.g) 

c, Do BD vuong AC => BD la duong cao 

CE vuong BA => CE la duong cao 

ma BD giao CE = I => I la truc tam 

=> AI la duong cao thu 3 

=> AI vuong BC 

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)