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a) \(x^2-\frac{2}{5}x< 0\)
\(\Leftrightarrow x\left(x-\frac{2}{5}\right)< 0\)
\(\Leftrightarrow\begin{cases}x>0\\x< \frac{2}{5}\end{cases}\) hoặc \(\begin{cases}x< 0\\x>\frac{2}{5}\end{cases}\) (loại)
\(\Leftrightarrow0< x< \frac{2}{5}\)
b) \(\frac{x-2}{x-6}< 0\)
\(\Leftrightarrow\begin{cases}x>2\\x< 6\end{cases}\) hoặc \(\begin{cases}x< 2\\x>6\end{cases}\) (loại)
\(\Leftrightarrow2< x< 6\)
c) \(\frac{x^2-1}{22}< 0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}x>1\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< 1\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< 1\)
a, \(\dfrac{20}{x}=\dfrac{-12}{15}\Rightarrow x=\dfrac{20.15}{-12}\Rightarrow x=-25\)
\(b,\dfrac{-15}{35}=\dfrac{27}{x}\Rightarrow x=\dfrac{35.27}{-15}\Rightarrow x=-63\)
\(c,\dfrac{\dfrac{4}{5}}{1\dfrac{2}{5}}=\dfrac{2\dfrac{2}{5}}{x}\Rightarrow x=\dfrac{2\dfrac{2}{5}.1\dfrac{2}{5}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{\dfrac{84}{25}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{21}{5}\)
\(d,\dfrac{x}{1\dfrac{1}{4}}=\dfrac{1\dfrac{1}{5}}{2}\Rightarrow x=\dfrac{1\dfrac{1}{4}.1\dfrac{1}{5}}{2}\Rightarrow x=\dfrac{\dfrac{3}{2}}{2}\Rightarrow x=\dfrac{3}{4}\)
\(e,\dfrac{\dfrac{1}{2}}{1\dfrac{1}{4}}=\dfrac{x}{3\dfrac{1}{3}}\Rightarrow x=\dfrac{\dfrac{1}{2}.3\dfrac{1}{3}}{1\dfrac{1}{4}}\Rightarrow x=\dfrac{\dfrac{5}{3}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{4}{3}\)
a) \(x^2+4x>0\)
\(\Leftrightarrow x\left(x+4\right)>0\)
\(\Leftrightarrow\begin{cases}x>0\\x>-4\end{cases}\) hoặc \(\begin{cases}x< 0\\x< -4\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>0\\x< -4\end{array}\right.\)
b) \(2\left(x-3\right)\left(x+7\right)>0\)
\(\Leftrightarrow\begin{cases}x>3\\x>-7\end{cases}\) hoặc \(\begin{cases}x< 3\\x< -7\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>3\\x< -7\end{array}\right.\)
c) \(\left(\frac{1}{2}-x\right)\left(\frac{1}{3}-x\right)>0\)
\(\Leftrightarrow\begin{cases}\frac{1}{2}-x>0\\\frac{1}{3}-x>0\end{cases}\) hoặc \(\begin{cases}\frac{1}{2}-x< 0\\\frac{1}{3}-x< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x< \frac{1}{2}\\x< \frac{1}{3}\end{cases}\) hoặc \(\begin{cases}x>\frac{1}{2}\\x>\frac{1}{3}\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x< \frac{1}{3}\\x>\frac{1}{2}\end{array}\right.\)
=\(2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)
\(=2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right).....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)
a:
\(\Leftrightarrow x^2-25⋮x^2-4\)
\(\Leftrightarrow x^2-4\inƯ\left(21\right)\)
\(\Leftrightarrow x^2-4\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{1;-1;5;-5\right\}\)
b:
1: Để A là số nguyên thì \(x^2-x⋮x+1\)
\(\Leftrightarrow x^2+x-2x-2+2⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{0;-2;1;-3\right\}\)
2: Để B là số nguyên thì \(-x\left(x-2\right)-5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
Đề bài của bn có j sai sai
Nếu sai thật thì vt lại ik mk giải cho
Còn nếu thế là đúng thì mk chịu