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Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
$\textbf{a)}$
$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow14x-18=5+2x-14$
$\Leftrightarrow12x=9$
$\Leftrightarrow x=\dfrac34.$
$\textbf{b)}$
$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$
$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$
$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$
$\Leftrightarrow21(2x-3)=30$
$\Leftrightarrow42x=93$
$\Leftrightarrow x=\dfrac{31}{14}.$




