a: =>4^x=640

=>\(x\in\varnothing\)

b: =>\(3^{-2x}\cdot3^{3x}=243\)

=>3^x=243

=>x=5

Lời giải:

a) 

$3^{2x+1}.7^y=9.21^x=3^2.(3.7)^x=3^{2+x}.7^x$

Vì $x,y$ là số tự nhiên nên suy ra $2x+1=2+x$ và $y=x$

$\Rightarrow x=y=1$

b) \(\frac{27^x}{3^{2x-y}}=\frac{3^{3x}}{3^{2x-y}}=3^{x+y}=243=3^5\Rightarrow x+y=5(1)\)

\(\frac{25^x}{5^{x+y}}=\frac{5^{2x}}{5^{x+y}}=5^{x-y}=125=5^3\Rightarrow x-y=3\) $(2)$

Từ $(1);(2)\Rightarrow x=4; y=1$

\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)

2 điểm!?

thi hay sao?

S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )

S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

Đặt A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

1093/729

Úi

2A = 3 - \(\dfrac{1}{729}=\dfrac{2186}{729}\) 

A = \(\dfrac{2186}{729}:2=\dfrac{1093}{729}\)

`a)x^3=343=7^3`

`=>x=7`

Vậy `x=7`

`b)(x-2,5)^4=(x-2,5)^2`

`=>(x-2,5)^2[(x-2,5)^2-1]=0`

`+)(x-2,5)^2=0<=>x=2,5`

`+)(x-2,5)^2=1`

`TH1:x-2,5=1<=>x=3,5`

`th2:x-2,5=-1<=>x=1,5`

Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5

`c)x^8/243=27`

`=>x^8=27.243`

`=>x^8=3^3*3^5=3^8`

`=>x=+-3`

Giải:

a) \(x^3=343\) 

\(\Rightarrow x^3=7^3\) 

\(\Rightarrow x=7\) 

b) \(\left(x-2,5\right)^4=\left(x-2,5\right)^2\) 

Vì có cùng cơ số nhưng lại khác mũ thì \(\left(x-2,5\right)\in\left\{0;1\right\}\) 

TH1:

\(\Rightarrow x-2,5=0\) 

\(\Rightarrow x=2,5\) 

TH2:

\(\Rightarrow\left[{}\begin{matrix}x-2,5=1\\x-2,5=-1\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=1,5\end{matrix}\right.\) 

c) \(\dfrac{x^8}{243}=27\) 

\(\Rightarrow x^8=3^3.3^5\) 

\(\Rightarrow x^8=3^8\) 

\(\Rightarrow x=\pm3\)

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)