x+y=xy => x=xy-y=y(x-1)

=>x:y=x:y(x-1)=x-1 mà x+y=x:y => x+y=x-1 => x+y=x+(-1) => y=-1

thay x=-1 vào x+y=xy => x+(-1)=x(-1)=>x-1=-x => 2x=1 => x=1/2

Vậy x=1/2 và y=-1

Dễ thấy góc BAa và góc ABc là 2 góc so le trong mà\(\widehat{BAa}=\widehat{ABc}=120^o\)

=>a//c

Ta có: \(\widehat{ABc}+\widehat{ABC}+\widehat{cBC}=360^o\Leftrightarrow120^o+80^o+\widehat{cBC}=360^o\)

<=>\(\widehat{cBC}=160^o\)

Góc cBC và góc BCb so le trong mà góc cBC=góc BCb=160^o

=>b//c

\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)=>

(x−1)²=(y−2)²=0

(x-1)²=0=>x-1=0=>x=1
(y-2)²=0=>y-2=0=>y=2

Vậy x=1 và y=2

Vì \(\left\{{}\begin{matrix}\left|x^2-4\right|\ge0\\\left|y^2-9\right|\ge0\end{matrix}\right.\)=>|x² - 4 | + | y² - 9 | \(\ge\) 0

Dấu "=" xảy ra khi |x² - 4 | = | y² - 9 | = 0

|x² - 4 |=0 => x²-4=0 => x²=4 => \(x=\pm2\)

| y² - 9 |=0 =>y²-9=0=>y²=9=>\(y=\pm3\)

Vậy \(x=\pm2\) và \(y=\pm3\)

mình làm sai rồi bạn, làm lại này:

\(\dfrac{7}{1.2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}\)

\(=\dfrac{7}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{48.49.50}\right)\)

\(=\dfrac{7}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\right)\)

\(=\dfrac{7}{2}\left(\dfrac{1}{2}-\dfrac{1}{49.50}\right)=\dfrac{7}{4}-\dfrac{1}{7.100}=\dfrac{306}{175}\)

\(\dfrac{7}{1.2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}=7\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)

\(=7\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\right)\)

\(=7\left(\dfrac{1}{1.2}-\dfrac{1}{49.50}\right)=\dfrac{7}{2}-\dfrac{1}{7.50}=\dfrac{1225}{350}-\dfrac{1}{350}=\dfrac{1224}{350}=\dfrac{612}{175}\)

không phân tích được

a)\(N=\left(\dfrac{1}{y-1}-\dfrac{y}{1-y^3}.\dfrac{y^2+y+1}{y+1}\right):\dfrac{1}{y^2-1}\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{y^3-1}.\dfrac{y^2+y+1}{y+1}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\dfrac{y^2+y+1}{y+1}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y-1\right).\left(y+1\right)\)

\(N=y+1+y\)

N=2y+1

b) Khi y=1/2 thì N=\(2.\dfrac{1}{2}+1=1+1=2\)

c) N dương <=> N=2y+1>0 <=> 2y>-1 <=> y>-1/2

\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{50}\right)=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{49}{50}=\dfrac{1}{50}\)

\(P=\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+6\right)^{\left(x+5\right)}}}}=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}=3^{2^{1^{13^{12}}}}\)

\(=3^{2^1}=3^2=9\)