\(A=3^x+3^{x+1}+...+3^{x+100}\)

=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)

=>\(2A=3^{x+101}-3^x\)

=>\(A=\dfrac{3^{x+101}-3^x}{2}\)

=>\(3^{x+101}-3^x=3^{105}-3^4\)

=>x=4

x=4

\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)

\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)

\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)

Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)

\(\Rightarrow x-3=0\Leftrightarrow x=3\)

3ˣ⁺¹ + 3ˣ⁺³ = 810

3ˣ⁺¹.(1 + 3²) = 810

3ˣ⁺¹.10 = 810

3ˣ⁺¹ = 810 : 10

3ˣ⁺¹ = 81

3ˣ⁺¹ = 3⁴

x + 1 = 4

x = 4 - 1

x = 3

\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

\(\left(x+1\right)^{x+3}=\left(x+1\right)^{x+1}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)

`x/2+x+x/3+x+x+x/4=5 3/4`

`=>3x+x/2+x/3+x/4=23/4`

`=>49/12x=23/4`

`=>x=69/49`

Vậy `x=69/49`

giúp mik với xin các bạn đó mik đnag cần gấp

Soạn đề bằng thanh công cụ cho rõ nha bạn