Tính độ dài MN biết vtAM=2/7vtAC và vtBN=-1/2vtBC

1/ Tinh ∆. Pt co 2 nghiem x1,x2 <=> ∆>=0.
Theo dinh ly Viet: S=x1+x2=-b/a=m+3.
Theo gt: |x1|=|x2| <=> ...

2/ \(\frac{\sin^2x-\cos^2x}{1+2\sin x.\cos x}\)

\(=\frac{\cos^2x\left(\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)}{\cos^2x\left(\frac{1}{\cos^2x}+\frac{2\sin x.\cos x}{\cos^2x}\right)}\)

\(=\frac{\tan^2x-1}{\tan^2x+1+2\tan x}\)

\(=\frac{\left(\tan x-1\right)\left(\tan x+1\right)}{\left(\tan x+1\right)^2}\)

\(=\frac{\tan x-1}{\tan x+1}\left(dpcm\right)\)

c/

• Ta có: \(\overrightarrow{BA}^2=\left(\overrightarrow{CA}-\overrightarrow{CB}\right)^2\)

\(\Leftrightarrow BA^2=CA^2-2\overrightarrow{CA}.\overrightarrow{CB}+CB^2\)

\(\Leftrightarrow\overrightarrow{CA}.\overrightarrow{CB}=\frac{CA^2+CB^2-BA^2}{2}=\frac{77}{2}\)

• \(\overrightarrow{MN}^2=\left(\overrightarrow{CN}-\overrightarrow{CM}\right)^2=\left(\frac{3}{2}\overrightarrow{CB}-\frac{5}{7}\overrightarrow{CA}\right)^2\)

\(\Leftrightarrow MN^2=\frac{9}{4}CB^2-\frac{15}{7}\overrightarrow{CA}.\overrightarrow{CB}+\frac{25}{49}CA^2\)

\(=\frac{9}{4}.64-\frac{15}{7}.\frac{77}{2}+\frac{25}{49}.49\)

\(=\frac{173}{2}\)

\(\Rightarrow MN=\sqrt{\frac{173}{2}}=\frac{\sqrt{346}}{2}\)