Question

[Gia Huy To]

Answer 1

[HT.Phong (9A5)]

\(a)3^{2x-1}+2\cdot9^{x-1}=405\\ =>3^{2x-1}+2\cdot\left(3^2\right)^{x-1}=405\\ =>3^{2x-1}+2\cdot3^{2x-2}=405\\ =>3^{2x-2}\cdot\left(3+2\right)=405\\ =>3^{2x-2}\cdot5=405\\ =>3^{2x-2}=\dfrac{405}{5}=81\\ =>3^{2x-2}=3^4\\ =>2x-2=4\\ =>2x=4+2=6\\ =>x=\dfrac{6}{2}\\ =>x=3\)

\(b)\left(\dfrac{1}{3}\right)^{x-1}+5\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\left(1+5\cdot\dfrac{1}{3^2}\right)=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}\cdot\dfrac{14}{9}=\dfrac{14}{729}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{14}{729}:\dfrac{14}{9}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\dfrac{9}{729}=\dfrac{1}{81}\\ =>\left(\dfrac{1}{3}\right)^{x-1}=\left(\dfrac{1}{3}\right)^4\\ =>x-1=4\\ =>x=1+4\\ =>x=5\)

\(c)\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{4}=-\dfrac{1}{4}\\ =>\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)=0\\ =>3x^3-\dfrac{8}{9}=0\\ =>3x^3=\dfrac{8}{9}\\ =>x^3=\dfrac{8}{9}:3=\dfrac{8}{27}\\ =>x^3=\left(\dfrac{2}{3}\right)^3\\ =>x=\dfrac{2}{3}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: \(3^{2x-1}+2\cdot9^{x-1}=405\)

=>\(\dfrac{3^{2x}}{3}+2\cdot3^{2x-2}=405\)

=>\(\dfrac{1}{3}\cdot3^{2x}+2\cdot3^{2x}\cdot\dfrac{1}{9}=405\)

=>\(3^{2x}\cdot\left(\dfrac{1}{3}+\dfrac{2}{9}\right)=405\)

=>\(3^{2x}\cdot\dfrac{5}{9}=405\)

=>\(3^{2x}=405:\dfrac{5}{9}=405\cdot\dfrac{9}{5}=81\cdot9=3^6\)

=>2x=6

=>x=3

b: \(\left(\dfrac{1}{3}\right)^{x-1}+5\cdot\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot3+5\cdot\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{3}=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x\cdot\left(3+\dfrac{5}{3}\right)=\dfrac{14}{9^3}\)

=>\(\left(\dfrac{1}{3}\right)^x=\dfrac{14}{3^6}:\dfrac{14}{3}=\dfrac{3}{3^6}=\dfrac{1}{3^5}\)

=>x=5

c: \(\dfrac{3}{5}\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)

=>\(\dfrac{9}{5}x^3-\dfrac{24}{45}-\dfrac{1}{2}\cdot\dfrac{1}{2}+\dfrac{1}{4}=0\)

=>\(\dfrac{9}{5}x^3=\dfrac{24}{45}=\dfrac{8}{15}\)

=>\(x^3=\dfrac{8}{15}:\dfrac{9}{5}=\dfrac{8}{15}\cdot\dfrac{5}{9}=\dfrac{40}{135}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

=>\(x=\dfrac{2}{3}\)

d: \(\dfrac{7}{x}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}+\dfrac{9}{45}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)

=>x=15

e: \(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{5}{31}\)

=>\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

=>\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}=\dfrac{1}{93}\)

=>2x+3=93

=>2x=90

=>x=45