ABC=125

abc = 125 nhé bạn 

làm bạn nhé

Có nghĩa là abc = 125

bc=25 vì 25 x 5 = 125

k cho mik nhe

Đề không rõ ràng. Bạn xem lại 

a=1,25

b=c=5

Lời giải:

\(ab=\frac{3}{5}; bc=\frac{4}{5}; ac=\frac{3}{4}\Rightarrow (abc)^2=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)

\(\Rightarrow abc=\pm \frac{3}{5}\)

Nếu $abc=\frac{3}{5}$ thì:

$c=\frac{3}{5}: \frac{3}{5}=1$

$a=\frac{3}{5}: \frac{4}{5}=\frac{3}{4}$

$b=\frac{3}{5}: \frac{3}{4}=\frac{4}{5}$

Nếu $abc=-\frac{3}{5}$ thì:

$c=-\frac{3}{5}: \frac{3}{5}=-1$

$a=-\frac{3}{5}: \frac{4}{5}=\frac{-3}{4}$

$b=-\frac{3}{5}: \frac{3}{4}=\frac{-4}{5}$

100.a+bc-bc=2.bc=> 100.a=2.bc=>50.a=bc=> a=1 => bc=50

abc=150

a) Ta có:

\(AC^2=13^2=169\)

\(AB^2+BC^2=5^2+12^2=25+144=169\)

\(\Rightarrow AB^2+BC^2=AC^2\)

\(\Rightarrow\Delta ABC\) vuông tại B (theo định lý Pytago đảo)

b) Ta có:

\(sinA=cosC=\dfrac{BC}{AC}=\dfrac{12}{13}\)

\(cosA=sinC=\dfrac{AB}{AC}=\dfrac{5}{13}\)

\(tanA=cotC=\dfrac{BC}{AB}=\dfrac{12}{5}\)

\(cotA=tanC=\dfrac{AB}{BC}=\dfrac{5}{12}\)

a. \(\Delta ABC\) có

\(AB^2+BC^2=5^2+12^2=169\)

\(AC^2=13^2=169\)

\(\Rightarrow AC^2=AB^2+BC^2\)

\(\Rightarrow\Delta ABC\perp tại.B\)

b. \(sin.A=\dfrac{BC}{AC}=\dfrac{12}{13}\\ cos.A=\dfrac{AB}{AC}=\dfrac{5}{13}\\ tan.A=\dfrac{BC}{AB}=\dfrac{12}{5}\\ cot.A=\dfrac{AB}{BC}=\dfrac{5}{12}\)

\(sin.C=\dfrac{AB}{AC}=\dfrac{5}{13}\\ cos.C=\dfrac{BC}{AC}=\dfrac{12}{13}\\ tan.C=\dfrac{AB}{BC}=\dfrac{5}{12}\\ cot.C=\dfrac{BC}{AB}=\dfrac{12}{5}\)