Question

Đề bài

Giải mỗi bất phương trình sau:

a)     \({3^x} > \frac{1}{{243}}\)

b)    \({\left( {\frac{2}{3}} \right)^{3x - 7}} \le \frac{3}{2}\)

c)     \({4^{x + 3}} \ge {32^x}\)

d)    \(\log (x - 1) < 0\)

e)     \({\log _{\frac{1}{5}}}(2x - 1) \ge {\log _{\frac{1}{5}}}(x + 3)\)

f)      \(\ln (x + 3) \ge \ln (2x - 8)\)

Answer 1

\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)

Answer 2

d, Điều kiện: x > 1

\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)

e, Điều kiện: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)

f, Điều kiện: x > 4

\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)