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X-1/x+2=x-2/x+3
=>(x-1)(x+3)=(x+2)(x-2)
=>x(x+3)-1(x+3)=x(x-2)+2(x-2)
=>x^2+3x-x-3=x^2-2x+2x-4
=>x^2+2x-3=x^2-4
=>2x-3=-4=>x=-1/2=-0,5
vậy...
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-x+3x-3=x^2-2x+2x-4\)
\(\Leftrightarrow x^2+2x-3=x^2-4\)
\(\Leftrightarrow x^2-x^2+2x=3-4\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
\(\frac{37-x}{x+13}=\frac{3}{7}\)
=>7.(37-x)=3.(x+13)
<=>259-7x=3x+39
<=>3x+7x=259-39
<=>10x=220
<=>x=22
\(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow\left(37-x\right).7=\left(x+13\right).3\Leftrightarrow259-7x=3x+39\)(nhân chéo)
\(\Leftrightarrow3x+7x=259-39\Rightarrow10x=220\Rightarrow x=220:10\Rightarrow x=22\)
Vậy x=22
\(\frac{-2}{3}\)\(.\)\(x\)\(=\)\(\frac{4}{5}\)
=> \(x\)\(=\)\(\frac{4}{5}\)\(:\)\(\frac{-2}{3}\)
\(x\)\(=\)\(\frac{4}{5}\)\(.\)\(\frac{-3}{2}\)
\(x\)\(=\)\(\frac{-6}{5}\)
Vậy đáp án C đúng
Bài 2:
TH1: \(x\le-\frac{5}{2}\)
<=>\(-\left(x+\frac{5}{2}\right)+\frac{2}{5}-x=0\)<=>\(-x-\frac{5}{2}+\frac{2}{5}-x=0\)<=>\(-\frac{21}{10}-2x=0\)
<=>\(-2x=\frac{21}{10}\)<=>\(x=\frac{-21}{20}\)(loại)
TH2: \(-\frac{5}{2}< x\le\frac{2}{5}\)
<=>\(x+\frac{5}{2}+\frac{2}{5}-x=0\)<=>\(\frac{29}{10}=0\)(loại)
TH3: \(x>\frac{2}{5}\)
<=>\(x+\frac{5}{2}+x-\frac{2}{5}=0\)<=>\(2x+\frac{21}{10}=0\)<=>\(2x=-\frac{21}{10}\)<=>\(x=-\frac{21}{20}\)(loại)
Vậy không có số x thỏa mãn đề bài
Bài 1:
Vì \(\left(x-2\right)^2\ge0\) nên\(\left(x-2\right)^2\le0\) khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Bài 3:
Đặt \(\frac{x}{15}=\frac{y}{9}=k\Rightarrow\hept{\begin{cases}x=15k\\y=9k\end{cases}}\)
Theo đề bài: xy=15 <=> 15k.9k=135k2=15 <=> k2=1/9 <=> k=-1/3 hoặc k=1/3
+) \(k=-\frac{1}{3}\Rightarrow\hept{\begin{cases}x=\left(-\frac{1}{3}\right).15=-5\\y=\left(-\frac{1}{3}\right).9=-3\end{cases}}\)
+) \(k=\frac{1}{3}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}.15=5\\y=\frac{1}{3}.9=3\end{cases}}\)
Vậy ...........
\(\frac{11}{14}+\left|\frac{2}{7}-x\right|-\frac{5}{2}=\frac{4}{3}\)
\(\Leftrightarrow\frac{11}{14}+\left|\frac{2}{7}-x\right|=\frac{23}{6}\)
\(\Leftrightarrow\left|\frac{2}{7}-x\right|=\frac{64}{21}\)
\(\Leftrightarrow\frac{2}{7}-x=\pm\frac{64}{21}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{2}{7}-x=\frac{64}{21}\\\frac{2}{7}-x=-\frac{64}{21}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{58}{21}\\x=\frac{10}{3}\end{array}\right.\)
Mà \(x>0\)
Vậy \(x=\frac{10}{3}\)