ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\end{matrix}\right.\)

PT trở thành:

\(\dfrac{x}{x-2}=\dfrac{1}{x+1}+\dfrac{x+1}{x+1}\\ \Leftrightarrow\dfrac{x}{x-2}=\dfrac{x+2}{x+1}\\ \Leftrightarrow x\left(x+1\right)=x^2-4\\ \Leftrightarrow x^2+x-x^2+4=0\\ \Leftrightarrow x+4=0\Leftrightarrow x=-4\left(tm\right)\)

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

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

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

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

a: Ta có: \(AM=MB=\dfrac{AB}{2}\)

\(DN=NC=\dfrac{DC}{2}\)

\(BE=EC=\dfrac{BC}{2}\)

mà AB=DC=BC

nên AM=MB=DN=NC=BE=EC

Xét tứ giác AMCN có

AM//CN

AM=CN

Do đó: AMCN là hình bình hành

b: Xét ΔMBC vuông tại B và ΔECD vuông tại C có

MB=EC

BC=CD

Do đó: ΔMBC=ΔECD

=>\(\widehat{BMC}=\widehat{CED}\)

=>\(\widehat{CED}+\widehat{ECM}=90^0\)

=>CM\(\perp\)DE

c: ΔMBC=ΔECD

=>MC=ED

a: ĐKXĐ: \(x\notin\left\{-1;0;1\right\}\)

\(\dfrac{x+3}{x+1}-\dfrac{x-1}{x}=\dfrac{3x^2+4x+1}{x\left(x-1\right)}\)

=>\(\dfrac{x\left(x+3\right)-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{3x^2+4x+1}{x\left(x-1\right)}\)

=>\(\dfrac{x^2+3x-x^2+1}{\left(x^2+x\right)}=\dfrac{3x^2+4x+1}{x\left(x-1\right)}\)

=>\(\dfrac{\left(3x+1\right)\left(x-1\right)}{x\left(x+1\right)\left(x-1\right)}=\dfrac{\left(3x^2+4x+1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}\)

=>\(\left(3x^2+4x+1\right)\left(x+1\right)=\left(3x+1\right)\left(x-1\right)\)

=>\(\left(3x+1\right)\left(x^2+2x+1\right)-\left(3x+1\right)\left(x-1\right)=0\)

=>\(\left(3x+1\right)\left(x^2+2x+1-x+1\right)=0\)

=>\(\left(3x+1\right)\left(x^2+x+2\right)=0\)

mà \(x^2+x+2=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)

nên 3x+1=0

=>\(x=-\dfrac{1}{3}\left(nhận\right)\)

b: ĐKXĐ: \(x\notin\left\{-3;-1\right\}\)

\(\dfrac{x}{2\left(x+3\right)}+\dfrac{x}{2x+2}=\dfrac{-x}{\left(x+1\right)\left(x+3\right)}\)

=>\(\dfrac{x}{2\left(x+3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{-x}{\left(x+1\right)\left(x+3\right)}\)

=>\(\dfrac{x\left(x+1\right)+x\left(x+3\right)}{2\left(x+3\right)\left(x+1\right)}=\dfrac{-2x}{2\left(x+1\right)\left(x+3\right)}\)

=>\(x^2+x+x^2+3x=-2x\)

=>\(2x^2+6x=0\)

=>2x(x+3)=0

=>x(x+3)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{0;\dfrac{3}{2}\right\}\)

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

=>\(\dfrac{x-3}{x\left(2x-3\right)}=\dfrac{x}{5}\)

=>\(x^2\left(2x-3\right)=5\left(x-3\right)\)

=>\(2x^3-3x^2-5x+15=0\)

=>\(x\simeq-1,9\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)

=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)

=>\(\left(x+2\right)^2=x^2+5x+4+x^2\)

=>\(2x^2+5x+4-x^2-4x-4=0\)

=>\(x^2+x=0\)

=>x(x+1)=0

=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Bạn viết rõ lại đề nhé ! 

\(x^4+1997x^2+1996x+1997\)

\(=\left(x^4+x^3+x^2\right)+\left(-x^3-x^2-x\right)+\left(1997x^2+1997x+1997\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

4(x-2)-3(x+1)=5

=>\(4x-8-3x-3=5\)

=>\(x-11=5\)

=>x=11+5=16

\(4\left(x-2\right)-3\left(x+1\right)=5\)

\(\Leftrightarrow4x-8-3x-3=5\)

\(\Leftrightarrow\left(4x-3x\right)=5+8+3\)

\(\Leftrightarrow x=16\)

Vậy \(x=16\)

\(sin^210+sin^220+sin^245+sin^270+sin^280\)

\(=sin^210+sin^220+sin^245+cos^220+cos^210=1+1+sin^245=2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(sin^210^0+sin^220^0+sin^245^0+sin^270^0+sin^280^0\)

\(=\left(sin^210^0+sin^280^0\right)+\left(sin^220^0+sin^270^0\right)+\left(sin^245^0\right)\)

\(=\left(sin^210^0+cos^210^0\right)+\left(sin^220^0+cos^220^0\right)+\left(\dfrac{1}{2}\right)\)

\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)

\(x^3+5x^2+8x+4\)

\(=x^3+x^2+4x^2+4x+4x+4\)

\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+4x+4\right)\)

\(=\left(x+1\right)\left(x+2\right)^2\)