Ta có : \(A=\dfrac{x^2}{x+1}=\dfrac{x^2+2x+1-2x-1}{x+1}=\dfrac{\left(x+1\right)^2-2x-2+1}{x+1}\)

\(=\dfrac{\left(x+1\right)^2-2\left(x+1\right)+1}{x+1}=x+1-2+\dfrac{1}{x+1}=x-1+\dfrac{1}{x+1}\)

- Để A là số nguyên .

\(\Leftrightarrow x+1\inƯ_{\left(1\right)}\)

\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)

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

Vậy ...

\(\frac{5}{n+1}=\frac{n+1}{5}\)

\(\Leftrightarrow\left(n+1\right)^2=5^2\)

\(\Leftrightarrow\sqrt{\left(n+1\right)^2}=\sqrt{5^2}\)

\(\Leftrightarrow n+1=5\)

\(\Leftrightarrow n=5-1\)

\(\Leftrightarrow n=4\)

\(\frac{2x+1}{5}=\frac{x+2}{8}\)

\(\Leftrightarrow8\left(2x+1\right)=5\left(x+2\right)\)

\(\Leftrightarrow16x+8=5x+10\)

\(\Leftrightarrow11x=2\)

\(\Leftrightarrow x=\frac{2}{11}\)

Ta có : \(\frac{x+1}{5}=\frac{2x-7}{3}\)

\(\Rightarrow3\left(x+1\right)=5\left(2x-7\right)\)

\(\Leftrightarrow3x+3=10x-35\)

\(\Leftrightarrow3x-10x=-35-3\)

\(\Leftrightarrow-7x=-38\)

\(\Rightarrow x=\frac{38}{7}\)

Ta có : \(\frac{x}{4}=\frac{9}{x}\)

\(\Rightarrow x^2=9.4\)

=> x² = 36

=> x = +4;-4 

Để 3n + 4 chia hết cho n - 2

=> 3n - 6 + 10 chia hết cho n - 2

=> 3(n - 2) + 10 chia hết cho n - 2

=> 10 chia hết cho n - 2

=> n - 2 thuộc Ư(10) = {-10;-5;-2;-1;1;2;5;10}

Ta có bảng : 

trả loi nhanh = t.i.c.k

Toán lớp 6 đây sao ???
 

Ta có:

A=x-y+z+1

B=x-2

Vì a=b

=>x-y+z+1=x-2

=>z-y=x-2-x-1

=>z-y=(x-x)-(2+1)

=>z-y=-3

Mik ko hỉu chỗ này.