Vì \(S.MNP\) là hình chóp tam giác đều 

nên \(SM=SN=SP=5\left(cm\right)\) và \(\triangle MNP\) đều (t/c)

\(\Rightarrow\left\{{}\begin{matrix}MN=NP=PM=10\left(cm\right)\\\widehat{MNP}=\widehat{NPM}=\widehat{PMN}=60^{\circ}\end{matrix}\right.\)

\(\dfrac{-8}{24}=\dfrac{-8:8}{24:8}=\dfrac{-1}{3}\)

\(\dfrac{-124}{-100}=\dfrac{124}{100}=\dfrac{124:4}{100:4}=\dfrac{31}{25}\)

cảm ơn nhé !

\(2^3+2=8+2=10\)

\(3^2+3^3=9+27=36\)

\(2^3+2=8+2=10\)

\(3^2+3^3=9+27=36\)

\(\dfrac{5}{6}=\dfrac{20}{24}=\dfrac{35}{42}=\dfrac{40}{48}\)

Ta có:

\(\dfrac{5}{6}=\dfrac{20}{24}=\dfrac{35}{42}=\dfrac{40}{48}\)

Vậy số cần điền vào các chỗ trống lần lượt là 20; 42 và 40.

ĐKXĐ: \(x\ne2\)

\(\dfrac{x^2}{2-x}+\dfrac{3x-1}{3}=\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=\dfrac{5}{3}-\dfrac{3x-1}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=\dfrac{6-3x}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=2-x\)

\(\Rightarrow x^2=\left(2-x\right)^2\)

\(\Leftrightarrow x^2=x^2-4x+4\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\) (tm ĐKXĐ)

20 số

20 số :(

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(HB=\dfrac{12^2}{16}=9\left(cm\right)\)

BC=BH+CH=9+16=25(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{9\cdot25}=15\left(cm\right)\\AC=\sqrt{16\cdot25}=20\left(cm\right)\end{matrix}\right.\)

ĐKXĐ: \(x\ne0;x\ne2\)

\(\dfrac{x+2}{x}=\dfrac{x-3}{x-2}\)

\(\Rightarrow\left(x+2\right)\left(x-2\right)=x\left(x-3\right)\)

\(\Leftrightarrow x^2-4=x^2-3x\)

\(\Leftrightarrow-3x=-4\)

\(\Leftrightarrow x=\dfrac{4}{3}\) (tm ĐKXĐ)

a) \(S_{EAG}=\dfrac{1}{2}\times AG\times ED=\dfrac{1}{2}\times2\times3=3\left(cm^2\right)\)

\(S_{PBC}=\dfrac{1}{2}\times BC\times DC=\dfrac{1}{2}\times5\times5=12,5\left(cm^2\right)\)

b) Ta có:

\(S_{EBC}=\dfrac{1}{2}\times BC\times EC=\dfrac{1}{2}\times5\times8=20\left(cm^2\right)\)

\(S_{PEC}=S_{ECB}-S_{PBC}=20-12,5=7,5\left(cm^2\right)\)

Vậy nên:

\(PD=\dfrac{2\times S_{PEC}}{EC}=\dfrac{2\times7,5}{8}=1,875\left(cm\right)\)

c) Ta thấy:

\(\dfrac{IM}{IP}=\dfrac{S_{MIG}}{S_{IPG}}=\dfrac{S_{MIE}}{S_{IPE}}\) nên \(\dfrac{IM}{IP}=\dfrac{S_{MGE}}{S_{GPE}}=\dfrac{\dfrac{1}{2}\times MG\times3}{\dfrac{1}{2}\times GP\times3}=\dfrac{MG}{GP}\)

Kéo dài AD cắt EF tại K.

Ta có \(S_{AKM}=\dfrac{1}{2}\times3\times2=3\left(cm^2\right)\)

nên \(S_{EKM}=S_{AKE}-S_{AKM}=\dfrac{1}{2}\times3\times5-3=4,5\left(cm^2\right)\)

Vậy \(FM=\dfrac{2\times S_{EKM}}{KE}=1,8\left(cm\right)\)

Thế thì \(MG=3-1,8=1,2\left(cm\right)\)

Lại có \(GP=3-1,875=1,125\left(cm\right)\)

Vậy nên:

\(\dfrac{IM}{IP}=\dfrac{MG}{GP}=\dfrac{1,2}{1,125}=\dfrac{16}{15}\).