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Câu 2 :
b) \(\frac{x}{3}=\frac{-2}{9}\)
=> x = \(\frac{-2}{9}.3\) = \(\frac{-2}{3}\)
c) \(0,5x-\frac{2}{3}x=\frac{7}{12}\)
=> \(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
=> \(-\frac{1}{6}\)x = \(\frac{7}{12}\)
=> x = \(\frac{7}{12}:\frac{-1}{6}\)
=> x =\(\frac{-7}{2}\)
Đề 1 câu 5 :
\(3B=3^2+3^3+3^4+...+3^{201}\)
\(\Rightarrow2B=3B-B=3^{201}-3\)
\(\Rightarrow2B+3=\left(3^{201}-3\right)+3=3^{201}\)
Do đó n = 201
a/ ĐKXĐ: ...
\(\Leftrightarrow\frac{x+1}{2\left(x^2-1\right)}+\frac{3}{x^2-1}=\frac{1}{4}\)
\(\Leftrightarrow\frac{x+7}{x^2-1}=\frac{1}{2}\Leftrightarrow2x+14=x^2-1\)
\(\Leftrightarrow x^2-2x-15=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\frac{x-1}{x}=a\)
\(a-\frac{3}{2a}=-\frac{5}{2}\Leftrightarrow2a^2+5a-3=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-3\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x-1}{x}=-3\\\frac{x-1}{x}=\frac{1}{2}\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x+3\right)-\left(2x-3\right)}{4x^2-9}=\frac{4x^2-9-\left(2x^2-x-4\right)}{4x^2-9}\)
\(\Leftrightarrow2x^2+5x+9=2x^2+x-5\)
\(\Leftrightarrow4x=-14\Rightarrow x=-\frac{7}{2}\)
a/ \(\frac{-25}{\left(-x+2\right)\left(-3x-2\right)}< 0\Leftrightarrow\left[{}\begin{matrix}x< -\frac{2}{3}\\x>2\end{matrix}\right.\)
b/ \(\frac{1}{x-1}-\frac{2}{2x-1}>0\Leftrightarrow\frac{1}{\left(x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)
c/ \(\frac{2}{3-x}+\frac{2}{x-3}\le0\Leftrightarrow0\le0\) (luôn đúng)
Vậy nghiệm của BPT là \(R\backslash\left\{3\right\}\)
d/ \(1-\frac{x-1}{x^2-3x+2}\ge\Leftrightarrow\frac{x^2-4x+3}{\left(x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\ge3\\1< x< 2\\x< 1\end{matrix}\right.\)
e/ \(\frac{x+1}{x^2+x+2}-\frac{1}{x+1}>0\Leftrightarrow\frac{x-1}{\left(x+1\right)\left(x^2+x+2\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
d.
\(\Leftrightarrow\frac{2-5x+2\left(x+5\right)}{\left(x+5\right)\left(2-5x\right)}\ge0\)
\(\Leftrightarrow\frac{3\left(4-x\right)}{\left(x+5\right)\left(2-5x\right)}\ge0\Rightarrow\left[{}\begin{matrix}-5< x< \frac{2}{5}\\x\ge4\end{matrix}\right.\)
e.
\(\Leftrightarrow\frac{1}{x+2}+\frac{1}{x-1}-\frac{1}{x-2}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)+\left(x+2\right)\left(x-2\right)-\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)\left(x-2\right)}\ge0\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{\left(x+2\right)\left(x-1\right)\left(x-2\right)}\ge0\Rightarrow\left[{}\begin{matrix}-2< x\le0\\1< x< 2\\x\ge4\end{matrix}\right.\)
f.
\(\Leftrightarrow\frac{x-2}{x-3}-\frac{x-1}{x+1}>0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+1\right)-\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}>0\Leftrightarrow\frac{3x-5}{\left(x+1\right)\left(x-3\right)}>0\) \(\Rightarrow\left[{}\begin{matrix}-1< x\le\frac{5}{3}\\x>3\end{matrix}\right.\)
sao khó thế