Question

bài 1: tìm x biết x \(\in\) N

a, \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\) \(\times\) \(\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}\) = 2^x

b, (x - 1)^x+2 = (x - 1)^x+4

c, 6. (6 - x)²⁰⁰³ = (6 - x)²⁰⁰³

d, 4¹¹ . 25¹¹ \(\le\) 2^x . 5^x < 20¹² . 5¹²

Answer 1

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)

\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)

=>x=10

b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)

hay \(x\in\left\{0;1;2\right\}\)

c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)

\(\Leftrightarrow6-x=0\)

hay x=6