Theo đề ta có :

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\left(x+17\right)-\left(x+2\right)=x\)

\(\Rightarrow x=15\)

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)$\left(\genfrac{}{}{0ex}{}{1}{2}+1,5\right)\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$2\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{1}{5}:2$

$x=\genfrac{}{}{0ex}{}{1}{10}$
b) $\left(-1\genfrac{}{}{0ex}{}{3}{5}+x\right):\genfrac{}{}{0ex}{}{12}{13}=2\genfrac{}{}{0ex}{}{1}{6}$

$-1\genfrac{}{}{0ex}{}{3}{5}+x=\genfrac{}{}{0ex}{}{13}{6}\cdot \genfrac{}{}{0ex}{}{12}{13}$
$x=2+1\genfrac{}{}{0ex}{}{3}{5}$

$x=3\genfrac{}{}{0ex}{}{3}{5}$
c) $\left(x:2\genfrac{}{}{0ex}{}{1}{3}\right)\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x\cdot \genfrac{}{}{0ex}{}{3}{7}\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x=\genfrac{}{}{0ex}{}{-3}{8}:\genfrac{}{}{0ex}{}{3}{49}$
$x=\genfrac{}{}{0ex}{}{-49}{8}=-6\genfrac{}{}{0ex}{}{1}{8}$
d) $\genfrac{}{}{0ex}{}{-4}{7}\cdot x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}:\left(-1\genfrac{}{}{0ex}{}{2}{3}\right)$

$\genfrac{}{}{0ex}{}{-4}{7}x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{-3}{5}$
$-\genfrac{}{}{0ex}{}{4}{7}x=\genfrac{}{}{0ex}{}{-3}{40}-\genfrac{}{}{0ex}{}{7}{5}x=\genfrac{}{}{0ex}{}{-59}{40}:\genfrac{}{}{0ex}{}{-4}{7}=\genfrac{}{}{0ex}{}{413}{160}=2\genfrac{}{}{0ex}{}{93}{160}$
 

Sửa:\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)

Vậy x = 15

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7

Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)

Tương tự:

\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)

\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)

\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)

Giải:

a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)

\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)

\(\Leftrightarrow x=-\dfrac{79}{210}\)

Vậy \(x=-\dfrac{79}{210}\).

b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)

\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)

\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)

\(\Leftrightarrow3x=\dfrac{67}{75}\)

\(\Leftrightarrow x=\dfrac{67}{75}:3\)

\(\Leftrightarrow x=\dfrac{67}{225}\)

Vậy \(x=\dfrac{67}{225}\).

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\(\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}\)\(=\dfrac{x}{15}\cdot\dfrac{15}{\left(x+2\right)\left(x+17\right)}\) \(\dfrac{1}{x+2}-\dfrac{1}{x+17}\)\(=\dfrac{x}{15}\cdot\left(\dfrac{1}{x+2}-\dfrac{1}{x+17}\right)\)

\(\dfrac{x}{15}=\left(\dfrac{1}{x+2}-\dfrac{1}{x+17}\right):\left(\dfrac{1}{x+2}-\dfrac{1}{x+17}\right)\)

\(\dfrac{x}{15}=1\)

\(x=15\cdot1\)

\(x=15\)

Đề HSG mà