có nghiệm thôi nha

để M xác định 

\(\Rightarrow\orbr{\begin{cases}y-1\ne0\\y+1\ne0\end{cases}}\Rightarrow\frac{y\ne1}{y\ne-1}.\)

\(b,M=\frac{1}{y-1}+\frac{y}{y+1}+\frac{2y^2}{y^2-1}\)

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

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

\(M=\frac{y+1}{y-1}\)

c, Để M nhận giá trị nguyên 

\(\Rightarrow y+1⋮y-1\)

\(\Leftrightarrow y-1+2⋮y-1\)

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

y = .... Tự tính 

\(a\frac{x^2-49}{x+5}:\left(x-7\right)\)

\(=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{\left(x-7\right)}\)

\(=\frac{x+7}{x+5}\)

\(b,\frac{2x+7}{x+2}-\frac{x+8}{2x+4}\)

\(=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{4x+14-x+8}{2\left(x+2\right)}\)

\(=\frac{3x+22}{2\left(x+2\right)}\)

a) \(\frac{x^2-49}{x+5}\div\left(x-7\right)=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{x-7}=\frac{x+7}{x+5}\)

b) \(\frac{2x+7}{x+2}-\frac{x+8}{2x+4}=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{\left(4x+14\right)-\left(x+8\right)}{2\left(x+2\right)}\)

\(=\frac{4x+14-x-8}{2\left(x+2\right)}=\frac{3x+6}{2\left(x+2\right)}=\frac{3\left(x+2\right)}{2\left(x+2\right)}=\frac{3}{2}\)

\(\frac{3x^2-8x+13}{x^2+5}=\frac{x^2+5+2x^2-8x+8}{x^2+5}=1+\frac{2\left(x^2-4x+4\right)}{x^2+5}=1+\frac{2\left(x-2\right)^2}{x^2+5}\ge1\)

Dấu \(=\)xảy ra khi \(x-2=0\Leftrightarrow x=2\).

Ta có : A =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}\)

< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

= \(1-\frac{1}{n}=\frac{n-1}{n}\)

Ta có 3(n - 1) = 3n - 3

2n = 2n

Mà 3n - 3 > 2n

=> 3(n - 1) > 2n

=> \(\frac{2}{3}>\frac{n-1}{n}\)(tính chất tỉ lệ thức)

<=> A < 2/3 (ĐPCM)

tìm x,y nguyên thỏa mãn:))))

đề bài đây mk quên