cho 2 biểu thức
a=1+2+22+....+2200
b= 22011-1
giúp mình ik
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\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)
\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)
\(\Rightarrow A=2^{2011}-1=B\)
\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)
\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)
\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
\(=\dfrac{2^{20}.3^5.5^5}{2^{10}.3^5.5^3}=2^{10}.5^2=1024.25=25600\)
tik pls
\(\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^3}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^3}=2^{10}\cdot5^2=1024\cdot25=25600\)
A = 1 + 21 + 22 + 23 + ... + 22023
2.A = 2 + 22 + 23 + 24 + ... + 22024
2A - A = 2 + 22 + 23 + 22023 - (1 + 2 + 22 + 23 + 24 + ... + 22023)
A = 2 + 22 + 23 + 22023 + 22024- 1 - 2 - 22 - 23 - 24 - ... - 22023
A = (2 - 2) + (22 - 22) + (23 - 23) + (24 - 24) + ...+(22024 - 1)
A = 0 + 0 + 0 + 0 + .... + 22024 - 1
A = 22024 - 1
a=2201