Bài 6:

a: \(\dfrac{3}{4}-x=1\)

=>\(x=\dfrac{3}{4}-1=-\dfrac{1}{4}\)

b: \(x+4=\dfrac{1}{5}\)

=>\(x=\dfrac{1}{5}-4\)

=>\(x=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}\)

c: \(x-\dfrac{1}{5}=2\)

=>\(x=2+\dfrac{1}{5}\)

=>\(x=\dfrac{11}{5}\)

d: \(\dfrac{x}{-5}=\dfrac{6}{-10}\)

=>\(\dfrac{x}{-5}=\dfrac{3}{-5}\)

=>x=3

Bài 7:

a: 2/3 của 27 là \(\dfrac{2}{3}\cdot27=18\)

b: 5/6 của 300 là \(300\cdot\dfrac{5}{6}=250\)

Bài 5:

e: \(\left(\dfrac{-2}{-5}+\dfrac{-5}{-6}\right)+\dfrac{4}{5}\)

\(=\left(\dfrac{2}{5}+\dfrac{5}{6}\right)+\dfrac{4}{5}\)

\(=\dfrac{6}{5}+\dfrac{5}{6}=\dfrac{61}{30}\)

f: \(\dfrac{5}{13}\cdot\dfrac{17}{26}+\dfrac{5}{13}\cdot\dfrac{9}{26}\)

\(=\dfrac{5}{13}\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)

\(=\dfrac{5}{13}\cdot\dfrac{26}{26}=\dfrac{5}{13}\)

ĐKXĐ: x<>-1

Để B là số nguyên tố thì \(\left\{{}\begin{matrix}2-3x⋮x+1\\B>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3x-3+5⋮x+1\\\dfrac{2-3x}{x+1}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5⋮x+1\\\dfrac{3x-2}{x+1}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\in\left\{1;-1;5;-5\right\}\\-1< x< \dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2;4;-6\right\}\\-1< x< \dfrac{2}{3}\end{matrix}\right.\Leftrightarrow x=0\)

Thay x=0 vào B, ta được:

\(B=\dfrac{2-3\cdot0}{0+1}=\dfrac{2}{1}=2\) là số nguyên tố

Vậy: x=0

a: ĐKXĐ: \(n^2-2n\ne0\)

=>\(n\left(n-2\right)\ne0\)

=>\(n\notin\left\{0;2\right\}\)

b: Thay n=-3 vào A, ta được:

\(A=\dfrac{3}{\left(-3\right)^2-2\cdot\left(-3\right)}=\dfrac{3}{15}=\dfrac{1}{5}\)

Thay n=5 vào A, ta được:

\(A=\dfrac{3}{5^2-2\cdot5}=\dfrac{3}{25-10}=\dfrac{3}{15}=\dfrac{1}{5}\)

Câu 1:

\(\dfrac{1}{11}< \dfrac{1}{10}\)

\(\dfrac{1}{12}< \dfrac{1}{10}\)

...

\(\dfrac{1}{20}< \dfrac{1}{10}\)

Do đó: \(B=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}< \dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}=1\)(1)

\(\dfrac{1}{11}>\dfrac{1}{20}\)

\(\dfrac{1}{12}>\dfrac{1}{20}\)

...

\(\dfrac{1}{20}=\dfrac{1}{20}\)

Do đó: \(B=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}=\dfrac{10}{20}=\dfrac{1}{2}\)

Suy ra: \(\dfrac{1}{2}< B< 1\)

a: Trên tia Ox, ta có: OM<ON

nên M nằm giữa O và N

=>OM+MN=ON

=>MN+2=6

=>MN=4(cm)

b: Vì OM<MN

nên M không là trung điểm của ON

c: Elà trung điểm của MN

=>\(EM=EN=\dfrac{MN}{2}=2\left(cm\right)\)

Vì MO và ME là hai tia đối nhau

nên M nằm giữa O và E

=>EO=MO+ME=2+2=4(cm)

\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)

Do đó: \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}=\dfrac{1}{8}-\dfrac{1}{9}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=\dfrac{8}{9}\)

Suy ra: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\)

=>\(2A=1-\dfrac{1}{3^{99}}\)

=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{99}}< \dfrac{1}{2}\)