\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{12}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)

\(=\dfrac{1}{2}\)

Bài 1:

Để n thỏa mãn đ/k: \(25< 3^n< 250\) thì;

\(n\in\left\{3;4;5\right\}\)

Bài 2:

Gọi x là số tự nhiên cần tìm.

Theo đề bài ta có:

\(a=60q+31\left(q\in N\right)\)

\(a=12.17+r\left(0\le r< 12\right)\)

Ta lại có: \(60q⋮12\) và \(31:12\) dư \(7\)

\(\Rightarrow\) \(r=7\)

\(\Rightarrow\) \(a=12.17+7\)

\(a=211\)

Vậy \(a=211\)

\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{198}{199}=\dfrac{1}{199}\)

\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}=\dfrac{1}{100}\)

\(a,x-6=11\)

\(x=11+6\)

\(\Rightarrow x=17\)

\(b,\dfrac{254.399-145}{254+399.253}=\dfrac{\left(253+1\right).399-145}{254+399.253}\)

\(=\dfrac{253.399+399-145}{254+399.253}\)

\(=\dfrac{253.399+254}{254+399.253}\)

\(=1\)

Vì \(\dfrac{2x+1}{9}=\dfrac{-4}{18}\)

\(\Rightarrow\left(2x+1\right).18=-4.9\)

\(\left(2x+1\right).18=-36\)

\(2x+1=-36:18\)

\(2x+1=-2\)

\(2x=-2-1\)

\(2x=-3\)

\(\Rightarrow x=-3:2=\dfrac{-3}{2}\)

Vậy \(x=\dfrac{-3}{2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2.\left(x+y+z\right)}=\frac{1}{2}\)

=>x+y+z=1/2

2x=y+z+1=>3x=x+y+z+1=1/2+1=3/2

=>x=3/2:3=1/2

2y=z+x+1=>3y=x+y+z+1=1/2+1=3/2

=>y=3/2:3=1/2

2z=x+y-2=>3z=x+y+z-2=1/2-2=-3/2

=>x=-3/2:3=-1/2

Vậy\(x=\frac{1}{2},y=\frac{1}{2},z=-\frac{1}{2}\)

Ta có:

\(A=\dfrac{2011}{2014}=\dfrac{2014-3}{2014}=\dfrac{2014}{2014}-\dfrac{3}{2014}=1-\dfrac{3}{2014}\)

\(B=\dfrac{2014}{2017}=\dfrac{2017-3}{2017}=\dfrac{2017}{2017}-\dfrac{3}{2017}=1-\dfrac{3}{2017}\)

Vì \(\dfrac{3}{2014}>\dfrac{3}{2017}\Rightarrow1-\dfrac{3}{2014}>1-\dfrac{3}{2017}\)

\(\Leftrightarrow\dfrac{2011}{2014}>\dfrac{2014}{2017}\) Hay \(A>B\)

Vậy A>B