I DON NO

1. Ta có \(\frac{cosa+sina}{cosa-sina}=\frac{1+\frac{sina}{cosa}}{1-\frac{sina}{cosa}}=\frac{1+0,5}{1-0,5}=3.\)

2. Giả sử MN = 3a, MP = 4a, khi đó ta có: \(\frac{1}{9a^2}+\frac{1}{16a^2}=\frac{1}{12^2}\Rightarrow a=5\Rightarrow\hept{\begin{cases}MN=15\\MP=20\end{cases}}\)

Áp dụng hệ thức lượng trong tam giác vuông, ta có: \(NP=\sqrt{MN^2+MP^2}=25\left(cm\right)\)

bạn chỉ giúp mình câu b đii

b)

( Tu ve hinh nha! )

Xet AB < AC

Ta co : \(\Delta KBM\sim\Delta ICM\left(g-g\right)\)

\(\Rightarrow\dfrac{BK}{IC}=\dfrac{MK}{IM}\) \(\Rightarrow\dfrac{IC}{MI}=\dfrac{BK}{MK}\)

Co : goc BCM = goc BAM ( cung chan cung BM )

\(\Rightarrow\Delta AKM\sim\Delta CHM\)

\(\Rightarrow\dfrac{AK}{CH}=\dfrac{MK}{HM}\) \(\Rightarrow\dfrac{AK}{MK}=\dfrac{CH}{HM}\left(1\right)\)

Co : goc MBC = goc MAC ( cung chan cung MC )

=> \(\Delta AIM\sim\Delta BHM\left(g-g\right)\)

\(\Rightarrow\dfrac{AI}{BH}=\dfrac{IM}{HM}\) \(\Rightarrow\dfrac{AI}{IM}=\dfrac{BH}{HM}\left(2\right)\)

tu (1) va (2) => \(\dfrac{AK}{MK}+\dfrac{AI}{IM}=\dfrac{BH}{HM}+\dfrac{CH}{HM}\left(=\dfrac{BC}{HM}\right)\)

Lai co :

\(\dfrac{AB}{MK}+\dfrac{AC}{MI}=\dfrac{AK}{MK}-\dfrac{BK}{MK}+\dfrac{AI}{MI}+\dfrac{IC}{MI}=\dfrac{AK}{MK}+\dfrac{AI}{MI}\)( vi \(\dfrac{IC}{MI}=\dfrac{BK}{MK}\)) .

=> DPCM.