M xác định

\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

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

Thay x=5 ta có: 

\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)

Vậy \(M=5\)tại  x=5

\(M=0\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)

Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)

\(M=-1\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy với \(x=-1\)thì \(M=-1\)

\(M=\frac{4x+8}{x^2-1}:\frac{x+2}{x+1}-\frac{x-2}{1-x}\)   \(ĐKXĐ:x\ne\pm1\)

\(M=\frac{4\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{x+2}+\frac{x-2}{x-1}\)

\(M=\frac{4}{x-1}+\frac{x-2}{x-1}\)

\(M=\frac{4+x-2}{x-1}\)

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

vậy \(M=\frac{x+2}{x-1}\)

bạn ktra lại đề ở chỗ 2/3/-x 

Bài 1:

a) x≠2x≠2

Bài 2:

a) x≠0;x≠5x≠0;x≠5

b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x

c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.

=> x=−5x=−5

Bài 3:

a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)

=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5

=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5

=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5

=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25

=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25

=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25

=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x

=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x

=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x

=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x

=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x

c) tự làm, đkxđ: x≠1;x≠−1

ê k bn với mk ik

😘 😘 😘 😘

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

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

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)