Ta có: \(A=1+3^1+3^2+3^3+...+3^{199}+3^{200}\)

\(\Rightarrow3A=3^1+3^2+3^3+3^4+...+3^{201}\)

\(\Rightarrow3A-A=\left(3^1+3^2+3^3+3^4+...+3^{201}\right)-\left(1+3^1+3^2+3^3+...+3^{200}\right)\)

\(\Rightarrow2A=3^{201}-1\)

\(\Rightarrow A=\frac{3^{201}-1}{2}< 3^{201}-1< 3^{201}=B\)

Vậy A < B

Ta có : A = 1 + 3 + 3² + ... + 3²⁰⁰

\(\Leftrightarrow\)2A = 3 + 3² + 3³ + ... + 3²⁰¹

Lấy 2A - A = ( 3 + 3² + 3³ + ... + 3²⁰¹ ) - ( 1 + 3 + 3² + ... + 3²⁰⁰ )

\(\Rightarrow\)A = 3²⁰¹ - 1

Ta thấy : 3²⁰¹ - 1 < 3²⁰¹

\(\Leftrightarrow\)A < B

a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)

Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\); \(\left|y+4\right|\ge0\)\(\forall y\inℝ\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5 

gọi (d) y=x

Thay x=1=>y=1=> (1;1)

Thay x=2=>y=2=> (2;2)

gọi (d1) y=-2x

Thay x=-1=> y=2=> (-1;2)

Thay x=1=>y=-2=> (1;-2)

Đề thi ở tỉnh nào thế bạn

Bình Dương bạn nhé!

Huong dẫn: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) ( n\(\in\)N^*) áp dụng vào từng cái ngoặc 

\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{20}\left(1+2+3+......+20\right)\)

  \(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+......+\frac{1}{20}.\frac{20.21}{2}\)

   \(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+......+\frac{21}{2}=\frac{2+3+4+.....+21}{2}=\frac{20.23}{2}=230\)

Ủa dễ mà !

a) C = 4x

b) C= 3,14 . 2 . r

Bạn ơi sao lại là 4x và 3,14.2.r