Ta có : \(xy-x-y-1=0\)

\(\Rightarrow\left(xy-x\right)-y-1=0\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)-2=0\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=2\)

\(\Rightarrow\left(y-1\right)\left(x-1\right)=2\)

\(\Rightarrow y-1;x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng sau :

Vậy \(\left(x;y\right)\in\left\{\left(3;2\right),\left(2;3\right),\left(-1;0\right),\left(0;-1\right)\right\}\)  

Xy-x-y-5=4 nhanh giùm mìnhĐang cần gấp

a) Ta có

+)AM=AB-BM=6-3,75=2,25

+)MN//BC => \(\frac{AN}{AC}=\frac{AM}{AB}\)=> \(\frac{AN}{8}=\frac{2,25}{6}=\frac{3}{8}\)

=> AN=3(cm)

CN=AC-AN=8-3=5(cm)

b) +)MK//BI => \(\frac{MK}{BI}=\frac{AK}{AI}\left(1\right)\)

+) NK//CI => \(\frac{NK}{CI}=\frac{AK}{AI}\left(2\right)\)

(1)(2) => \(\frac{MK}{BI}=\frac{NK}{CI}\)mà MK=NK (K là trung điểm MN)

=> BI=CI => I là trung điểm BC

c) \(\Delta\)ABC vuông tại A

=> BC²=AB²+AC²=6²+8²=10² (Định lý Pytago)

=> BC=10cm

Ta có: \(\hept{\begin{cases}\frac{AN}{CN}=\frac{3}{5}\\\frac{AB}{BC}=\frac{6}{10}=\frac{3}{5}\end{cases}\Rightarrow\frac{AN}{CN}=\frac{AB}{AC}=\frac{3}{5}}\)

=> BN là phân giác \(\widehat{ABC}\)

Giải các pt sau:

a) (x+4)(2x-3)=0
TH1: x+4=0 => x=-4
TH2 : 2x-3=0 => 2x=3 =>x=3/2

b.

3x-1=7-x
=>3x-1-(7-x)=0
=>3x-1-7+x=0
=>4x-8=0
=>4x=8
=>x=2

a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)

ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)

b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)

\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)

c) \(P=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{\left(x+2\right)\left(2x+1-2\right)}{2x+1}=\frac{\left(x+2\right)\left(2x+1\right)-2\left(x+2\right)}{2x+1}\)

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

để P nguyền khi zà chỉ khi

\(1⋮2x+1\)

\(=>2x+1\inƯ\left(1\right)=\pm1\)

=>\(\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

hè lấu âu :>

ai chỉ với

Ta có:x^2-y^2-6y-9=x^2-(y^2+6y+9)

                              =x^2-(y+3)^2

                              =(x-y-3)(x-y+3)

Ta có : 

x² - y² + 6y - 9

= x² - ( y² + 6y + 9 )

= x² - ( y + 3 )²

= ( x - y - 3 ) ( x - y + 3 )

Ta có: \(\frac{\left(n-2\right).180}{n}=144\)

      \(\Rightarrow\left(n-2\right).180=144n\)

     \(\Leftrightarrow180n-360=144n\)

      \(\Rightarrow180n-144n=360\)

     \(\Leftrightarrow36n=360\)

      \(\Rightarrow n=360:36\)

      \(\Rightarrow n=10\)

      Vậy \(n=10\)

  Chúc bn hk tốt nha :)