\(B=m^2-mn+m-n^2-n+mn=m^2-n^2+n-n\\ =\left(m-n\right)\left(m+n+1\right)\\ =\left(-\dfrac{2}{3}+\dfrac{1}{3}\right)\left(-\dfrac{2}{3}-\dfrac{1}{3}+1\right)=-\dfrac{1}{3}\cdot0=0\)

Ta có: \(B=m\left(m-n+1\right)-n\left(n+1-m\right)\)

\(=m^2-mn+m-n^2-n+mn=m^2-n^2+m-n\)

=(m-n)(m+n)+(m-n)

=(m-n)(m+n+1)

\(=\left(-\frac23+\frac13\right)\left(-\frac23+\frac{-1}{3}+1\right)=0\)

Ta có: \(B=m\left(m-n+1\right)-n\left(n+1-m\right)\)

\(=m^2-mn+m-n^2-n+mn=m^2-n^2+m-n\)

=(m-n)(m+n)+(m-n)

=(m-n)(m+n+1)

\(=\left(-\frac23+\frac13\right)\left(-\frac23+\frac{-1}{3}+1\right)=0\)

\(B=m\left(m-n+1\right)-n\left(n+1-m\right)=m^2-mn+m-n^2-n+mn=m^2-n^2+m-n=\left(m-n\right)\left(m+n\right)+\left(m-n\right)=\left(m-n\right)\left(m+n+1\right)=\left(-\dfrac{2}{3}+\dfrac{1}{3}\right)\left(-\dfrac{2}{3}-\dfrac{1}{3}+1\right)=-\dfrac{1}{3}.0=0\)

Ta có: \(B=m\left(m-n+1\right)-n\left(n+1-m\right)\)

\(=m^2-mn+m-n^2-n+mn=m^2-n^2+m-n\)

=(m-n)(m+n)+(m-n)

=(m-n)(m+n+1)

\(=\left(-\frac23+\frac13\right)\left(-\frac23+\frac{-1}{3}+1\right)=0\)

a: Để C là số nguyên thì \(m^2-2m-m+2-5⋮m-2\)

\(\Leftrightarrow m-2\in\left\{1;-1;5;-5\right\}\)

hay \(m\in\left\{3;1;7;-3\right\}\)

c: Để E là số nguyên thì \(m+2⋮m^2-1\)

\(\Leftrightarrow m^2-1-3⋮m^2-1\)

\(\Leftrightarrow m^2-1\in\left\{1;-1;3;-3\right\}\)

hay \(m\in\left\{\sqrt{2};-\sqrt{2};0;2;-2\right\}\)

d: Để G là số nguyên thì \(3m+2⋮m^2-1\)

\(\Leftrightarrow9m^2-4⋮m^2-1\)

\(\Leftrightarrow m^2-1\in\left\{1;-1;5;-5\right\}\)

hay \(m\in\left\{\sqrt{2};-\sqrt{2};0;\sqrt{6};-\sqrt{6}\right\}\)

a: \(M=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{8\sqrt{x}}{1-x}\)

\(=\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(N=\frac{-x+\sqrt{x}-3}{x-1}-\frac{1}{\sqrt{x}-1}\)

\(=\frac{-x+\sqrt{x}-3-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

b: K=M:N

\(=\frac{-4\sqrt{x}}{x-1}\cdot\frac{x-1}{-x-4}=\frac{4\sqrt{x}}{x+4}\)

Thay \(x=14-6\sqrt5=\left(3-\sqrt5\right)^2\) vào K, ta được:

\(K=4\cdot\frac{\sqrt{\left(3-\sqrt5\right)^2}}{14-6\sqrt5+4}=\frac{4\left(3-\sqrt5\right)}{18-6\sqrt5}=\frac{4\left(3-\sqrt5\right)}{6\left(3-\sqrt5\right)}=\frac46=\frac23\)

Câu 2: 

2) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)

Câu 2 : 

Gọi : vận tốc của người đi chậm là : x (km/h) ( x > 0 ) 

Vận tốc của người đi nhanh : x + 4 (km/h) 

Vi : người đi chậm đến muộn hơn : 45 phút \(=\dfrac{3}{4}\left(h\right)\)

Khi đó : 

\(\dfrac{36}{x}-\dfrac{36}{x+4}=\dfrac{3}{4}\)

\(\Leftrightarrow\left[36\cdot\left(x+4\right)-36x\right]\cdot4=3x\cdot\left(x+4\right)\)

\(\Leftrightarrow3x^2+12x-144=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(n\right)\\x=16\left(l\right)\end{matrix}\right.\)

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)