Sửa đề: \(2^{x}+2^{x+1}+\cdots+2^{x+2021}=2^{2025}-8\)

Ta có: \(2^{x}+2^{x+1}+\cdots+2^{x+2021}=2^{2025}-8\)

=>\(2^{x}\left(1+2+2^2+\cdots+2^{2021}\right)=2^3\left(2^{2022}-1\right)\)

Đặt \(A=1+2+2^2+\cdots+2^{2021}\)

=>\(2A=2+2^2+2^3+\cdots+2^{2022}\)

=>\(2A-A=2+2^2+2^3+\cdots+2^{2022}-1-2-2^2-\cdots-2^{2021}\)

=>\(A=2^{2022}-1\)

TA có: \(2^{x}\left(1+2+2^2+\cdots+2^{2021}\right)=2^3\left(2^{2022}-1\right)\)

=>\(2^{x}\left(2^{2022}-1\right)=2^3\left(2^{2022}-1\right)\)

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

=>x=3

a: \(\left(x+2\right)^2+\left(x+8\right)\left(x+2\right)=0\)

=>(x+2)(x+2+x+8)=0

=>(x+2)(2x+10)=0

=>(x+2)(x+5)=0

=>\(\left[\begin{array}{l}x+2=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=-5\end{array}\right.\)

b: \(B=\left(x+y\right)\left(x^2-xy+y^2\right)-y^3\)

\(=x^3+y^3-y^3=x^3\)

Khi x=10 thì \(B=10^3=1000\)

1.     Giải:

Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)

 \(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)

 \(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)

Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.

⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)

Ta có bảng:

Vậy xϵ\(\left\{0;1;3;10\right\}.\)

2. Giải:

Do (2x-18).(3x+12)=0.

⇒ 2x-18=0             hoặc             3x+12=0.

⇒ 2x     =18                               3x       =-12.

⇒   x     =9                                   x       =-4.

Vậy xϵ\(\left\{-4;9\right\}.\)

3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.

S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.

S= 0 + 0 + ... + 0 + 2025.

⇒S= 2025.

a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\) 

   8 .x + 1 . x = 990

x . [ 8 +1 ] = 990

x . 9 = 990

x = 990 : 9

x = 110

các bạn giúp mình với mình đang vội.

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

ai giúp mik với

Ta có: \(A=\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot\left(1+\dfrac{2}{7}\right)\cdot...\cdot\left(1+\dfrac{2}{2021}\right)\)

\(=\dfrac{5}{3}\cdot\dfrac{7}{5}\cdot\dfrac{9}{7}\cdot...\cdot\dfrac{2023}{2021}\)

\(=\dfrac{2023}{3}\)

Sửa đề: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

Ta có: \(\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

\(=\left(1+\frac{2019}{2}\right)+\left(1+\frac{2018}{3}\right)+\cdots+\left(1+\frac{1}{2020}\right)+1\)

\(=\frac{2021}{2}+\frac{2021}{3}+\cdots+\frac{2021}{2021}=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

Ta có: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

=>x=2021