\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)

\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)

\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)

\(B=\dfrac{\left(x+4\right)\times x-2}{x+4}\)

\(B=x-\dfrac{2}{x+4}\)

Vì \(x\in z\), để \(B\in z\Leftrightarrow\dfrac{2}{x+4}\in z\)

                              \(\Leftrightarrow2⋮\left(x+4\right)\)

                              \(\Leftrightarrow x+4\inƯ\left(2\right)\)

Mà \(Ư\left(2\right)=\left(\pm1;\pm2\right)\)

Ta có bảng sau

\(\begin{matrix}x+4&1&-1&2&-2\\x&-3&-5&-2&-6\end{matrix}\)

Vậy \(x\in\left(-2;-3;-5;-6\right)\) thì \(B\in z\)

a)

Gọi x là số cần tìm, ta có:

 \(x+2>0\left(x>0\right)\)

\(\Rightarrow x-4< 0\)

\(\Rightarrow x< 4\)

\(x=\left\{1;2;3\right\}\)

b)

Gọi x là số cần tìm, khi đó:

\(x-2< 0\left(x< 0\right)\)

\(x+4>0\left(\forall x>-4\right)\)

\(\Rightarrow x=\left(-3;-2;-1\right)\)

Lời giải:

$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$

$=1$

`# \text {DNamNgV}`

\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}\)

\(\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)=\dfrac{1}{99}\)

\(\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}\)

\(\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}\)

\(\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}\)

\(\Rightarrow x=\dfrac{99}{99}\)

\(\Rightarrow x=1\)

Vậy, `x = 1.`

Lời giải:

$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12})=(x+1)(\frac{1}{13}+\frac{1}{14})$

$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14})=0$

Hiển nhiên $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$ 

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

Lời giải:

Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$

Đề bài 

$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow 12=x$ (thỏa mãn)

Trước tiên ta cần chứng minh : Với a<b thì : \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\) với c là số nguyên dương.

\(\Leftrightarrow a.\left(b+c\right)< b.\left(a+c\right)\)

\(\Leftrightarrow ab+ac< ab+bc\)

\(\Leftrightarrow ac< bc\)

\(\Leftrightarrow a< b\left(LĐ\right)\)

Áp dụng bổ đề đó ta có : \(\dfrac{a}{b+c}< \dfrac{a+a}{a+b+c}=\dfrac{2a}{a+b+c}\)

\(CMTT:\dfrac{b}{c+a}< \dfrac{2b}{a+b+c}\)

\(\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)

Do đó : \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< \dfrac{2a+2b+2c}{a+b+c}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\left(đpcm\right)\)

Lời giải:

Vì $b+c> a\Rightarrow 2(b+c)> a+b+c$

$\Rightarrow b+c> \frac{a+b+c}{2}$

$\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}(1)$

Hoàn toàn tương tự ta có:
$\frac{b}{c+a}< \frac{2b}{a+b+c}(2)$

$\frac{c}{a+b}< \frac{2c}{a+b+c}(3)$

Từ $(1); (2); (3)\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{2(a+b+c)}{a+b+c}=2$

Ta có đpcm.