Question

tìm x

a, \(\left(x+1\right)^5=3\)

b, \(\left(\dfrac{1}{2}-\dfrac{1}{6}\right)3^{x+4}-4.3^x=3^{17}-4.3^{13}\)

c, \(5^{x+4}-3.5^{x+3}=2.5^{11}\)

d, \(\left(\dfrac{1}{3}+\dfrac{1}{6}\right)2^{x+4}-2^x=2^{14}-2^{10}\)

e, \(\dfrac{3}{5}2^x+\dfrac{7}{5}2^{x+3}=\dfrac{3}{5}2^{10}+\dfrac{7}{5}2^{13}\)

 

Answer 1

a) 

\(\left(x+1\right)^5=3\\ =>x+1=\sqrt[5]{3}\\ =>x=\sqrt[5]{3}-1\) 

b) Xem lại đề 

c) 

\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\\ =>5^{x+3}\cdot\left(5-3\right)=2\cdot5^{11}\\ =>5^{x+3}\cdot2=2\cdot5^{11}\\ =>5^{x+3}=5^{11}\\ =>x+3=11\\ =>x=11-3\\ =>x=8\)
d) xem lại đề 

e) 

\(\dfrac{3}{5}\cdot2^x+\dfrac{7}{5}\cdot2^{x+3}=\dfrac{3}{5}\cdot2^{10}+\dfrac{7}{5}\cdot2^{13}\\ =>2^x\cdot\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\cdot\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\\ =>2^x=\dfrac{2^{10}\cdot\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)}{\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3}\\ =>2^x=2^{10}\\ =>x=10\)