Bạn đã ib nhờ mik thì mik làm cho trót vại UwU

\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-x\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}.\)

\(=-\frac{1}{x\left(x-y\right)\left(z-x\right)}-\frac{1}{y\left(y-z\right)\left(x-y\right)}-\frac{1}{z\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{y^2x-yz^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{xz^2-x^2z}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{x^2y-xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-y^2z+yz^2-xz^2+x^2z-x^2y+xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-\left(y^2z-x^2z\right)+\left(yz^2-xz^2\right)-\left(x^2y-xy^2\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-z\left(y^2-x^2\right)+z^2\left(y-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-z\left(y-x\right)\left(x+y\right)+z^2\left(y-x\right)+xy\left(y-x\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{-xyz\left(y-x\right)\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{-z\left(x+y\right)+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{-zx-zy+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-\left(zx-xy\right)-\left(zy-z^2\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-x\left(z-y\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{x\left(y-z\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-z\right)\left(x-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{x-z}{xyz\left(z-x\right)}\)

\(=\frac{-\left(z-x\right)}{xyz\left(z-x\right)}\)

\(=\frac{-1}{xyz}\)

b,

đổi dấu 

-(x-1)/2-x +1/2-x

=-x+1+1/2-x

=2-x/2-x

=1

Thặc vler .V

A/\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)

\(=\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\)

\(=\left[\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right]\)

\(=\left[\frac{x+3}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{x+5}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}+\frac{x+3}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\right]\)

\(=\frac{2x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2x+8}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)

\(=\frac{2\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2\left(x+4\right)}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)

\(=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}\)

\(=\frac{2x+10}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}+\frac{2x+2}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)

\(=\frac{4x+12}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)

\(=\frac{4\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)

\(=\frac{4}{\left(x+1\right)\left(x+5\right)}\)

B/\(\frac{x-1}{x-2}+\frac{1}{2-x}\)

\(=\frac{x-1}{x-2}-\frac{1}{x-2}\)

\(=\frac{x-1-1}{x-2}\)

\(=\frac{x-2}{x-2}\)

\(=1\)

\(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)

uầy

so amazing 

ai lm đc là legendary đó

viết cái đề nhìn zô đã ko mún làm r

`bạn tự kẻ hình nhé
ta đễ dàng cm dk DM=CM
Từ đó ta có SAMDAMD=1/2 SDACDAC=1/3 SABCABC
SBDMBDM = 1/2SBDCBDC= 1/6 SABCABC
Suy ra SABMABM=(1/3+1/6)SABCABC= 1/2SABCABC= 15m^2

cho hỏi là làm sao dm=cm

a) P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)

Vậy P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{\left(x-5\right)\left(x+5\right)2}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

Có: \(P=0\)

\(\Rightarrow P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=0\Leftrightarrow x\left(x^2+4x-5\right)=0\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow\left(x^2-x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)