a) Phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)

b) \(M=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{x+3}{x^2-1}\)

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

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

\(M=\frac{3x-3}{\left(x-1\right)\left(x+1\right)}\)

\(M=\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(M=\frac{3}{x+1}\)

Để M nguyên thì :

\(3⋮x+1\)

\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;-4\right\}\)( thỏa mãn ĐKXĐ )

Vậy.......

ĐKXĐ \(x\ne1;x\ne-1\)

\(M=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{x+3}{x^2-1}\)

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

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

\(M=\frac{3x-3}{\left(x-1\right)\left(x+1\right)}\)

\(M=\frac{3}{x+1}\)

Để M nguyên \(\Leftrightarrow\text{ }3\text{ }⋮\text{ }x+1\text{ }hay\text{ }x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

=>\(x\in\left\{-4;-2;0;2\right\}\)thì M nguyên

giúp mình với sắp thi rồi

\(x^2-4xy+5y^2+6x-10y+10=0\)

\(x^2-2x\left(2y-3\right)+5y^2-10y+10=0\)

\(x^2-2x\left(2y-3\right)+\left(4y^2-12x+9\right)+\left(y^2+2x+1\right)=0\)

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

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

Ta có: \(\hept{\begin{cases}\left(x-2y+3\right)^2\ge0\forall x;y\\\left(y+1\right)^2\ge0\forall y\end{cases}}\)\(\Rightarrow\left(x-2y+3\right)^2+\left(y+1\right)^2\ge0\forall x;y\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-2y+3\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2y+3=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-2y+3=0\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x+2+3=0\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)Vậy \(\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)

Tham khảo nhé~

Sao anh kudo không tách thẳng như vầy luôn cho nhanh?(nhanh hơn đúng 1 dòng ở phần phân tích thôi:v)

\(A=x^2-4xy+5y^2+6x-10y+10=0\)

\(\Leftrightarrow\left(x^2-2.x.2y+4y^2\right)+\left(6x-12y\right)+9+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left[\left(x-2y\right)^2+2.\left(x-2y\right).3+3^2\right]+\left(y+1\right)^2=0\)

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

Đến đây ez rồi!

Mình không chắc là có đúng không nữa các bạn xem hộ mình với nha!
= (100^2 - 99^2) + (98^2 - 97^2) + ... + (4^2 - 3^2) + (2^2 - 1^2) = 
= (100+99)(100-99) + (98+97)(98-97) + ... + (4+3)(4-3) + (2+1)(2-1) = 
= (100+99).1 + (98+97).1 + ... + (4+3).1 + (2+1).1 = 
= 100 + 99 + 98 + 97 + ... + 4 + 3 + 2 + 1 = 
= (100+1) + (99+2) + (98+3) + ... + (51+50) = 101.50 = 5050 
(50 cặp dấu ngoặc, tổng trong mỗi cặp dấu ngoặc là 101) 

Bài 1 :

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

\(\Rightarrow\left(2x+1\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=5\end{cases}}\)

KL :...

bài 2 :

\(x^2+6x+9-y^2=\left(x+3\right)^2-y^2\)

\(=\left(x+3-y\right)\left(x+3+y\right)\)

Bài 3 :

\(P=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\frac{3}{x-2}\)

\(\frac{x^3-2x^2+4}{x-2}\inℤ\Leftrightarrow x^3-2x^2+4⋮x-2\)

\(\Leftrightarrow x^3-2x^2-\left(x^3-2x^2\right)+4⋮x-2\Leftrightarrow4⋮x-2\)

\(\Leftrightarrow x-2\in\left\{-1;2;-2;1;-4;4\right\}\Leftrightarrow x\in\left\{1;4;0;3;-2;6\right\}\)

b, \(\frac{x^3-x^2+2}{x-1}\inℤ\Leftrightarrow x^3-x^2+2⋮x-1\)

\(\Leftrightarrow x^3-x^2-\left(x^3-x^2\right)+2⋮x-1\)

\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\in\left\{-1;1;-2;2\right\}\)

\(\Leftrightarrow x\in\left\{0;2;-1;3\right\}\)