\(\dfrac{7}{8}+x=\dfrac{7}{20}\)

=>\(x=\dfrac{7}{20}-\dfrac{7}{8}=\dfrac{14}{40}-\dfrac{35}{40}=-\dfrac{21}{40}\)

=)x=7/20-7/8

=)x=-21/40

\(\left(-1,5\right)\cdot\dfrac{-2}{3}+\left(2,5-\dfrac{3}{4}\right):1\dfrac{3}{4}\)

\(=\dfrac{-3}{2}\cdot\dfrac{-2}{3}+\left(2,5-0,75\right):1,75\)

=1+1

=2

CM:A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\) < 1

   \(\dfrac{1}{2^2}\) = \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}-\dfrac{1}{2}\)

   \(\dfrac{1}{3^2}\) = \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

    \(\dfrac{1}{4^2}\) = \(\dfrac{1}{4.4}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)

   \(\dfrac{1}{100^2}\) = \(\dfrac{1}{100.100}\) < \(\dfrac{1}{99.100}\) = \(\dfrac{1}{99}-\dfrac{1}{100}\)

Cộng vế với vế ta có:

    \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\)  = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\) < 1 (đpcm)

\(\dfrac{2}{3}x\) - \(\dfrac{1}{2}x\) = \(\dfrac{5}{12}\)

\(\dfrac{1}{6}x\)         = \(\dfrac{5}{12}\)

   \(x\)         = \(\dfrac{5}{12}\) : \(\dfrac{1}{6}\)

    \(x\)       = \(\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\) 

C = \(\dfrac{n+5}{n-3}\) (3 ≠ n; n \(\in\) Z)

C \(\in\) Z ⇔ n + 5 ⋮ n - 3

             n - 3 + 8 ⋮ n - 3

             n - 3 \(\in\) Ư(8) = {-8; -4; -2; -1; 1; 2; 4; 8}

Lập bảng ta có:

Theo bảng trên ta có n \(\in\) {-5; -1; 1; 2; 4; 5; 7; 11}

Vậy để C = \(\dfrac{n+5}{n-3}\) \(\in\) Z thì n \(\in\) {-5; -1; 1; 2; 4; 5; 7; 11}

ĐKXĐ: n<>3

Để C là số nguyên thì \(n+5⋮n-3\)

=>\(n-3+8⋮n-3\)

=>\(8⋮n-3\)

=>\(n-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(n\in\left\{4;2;5;1;7;-1;11;-5\right\}\)

\(60\%\cdot x+\dfrac{2}{3}x=-76\)

=>\(\dfrac{3}{5}x+\dfrac{2}{3}x=-76\)

=>\(x\left(\dfrac{3}{5}+\dfrac{2}{3}\right)=-76\)

=>\(x\cdot\dfrac{19}{15}=-76\)

=>\(x=-76:\dfrac{19}{15}=-76\cdot\dfrac{15}{19}=-60\)

ĐKXĐ: n<>3

Để B là số nguyên thì \(5⋮n-3\)

=>\(n-3\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{4;2;8;-2\right\}\)

Để B nguyên thì 5 ⋮ (n - 3)

⇒ n - 3 ∈ Ư(5) = {-5; -1; 1; 5}

⇒ n {-2; 2; 4; 8}