A= 2(x² -x + 1/4) - 1/2

GTNN = -1/2

k có đề bài toán nào cho kiểu này 19x -6x .....bởi z chán k mun làm

X=1 và C có zá chị = -5

\(C=\dfrac{20\sqrt{x}-16\sqrt{x}-8}{2\sqrt{x}+1}=\dfrac{20\sqrt{x}-8\left(2\sqrt{x}+1\right)}{2\sqrt{x}+1}=\dfrac{20\sqrt{x}}{2\sqrt{x}+1}-8\)

Do \(\left\{{}\begin{matrix}20\sqrt{x}\ge0\\2\sqrt{x}+1>0\end{matrix}\right.\) ; \(\forall x\Rightarrow\dfrac{20\sqrt{x}}{2\sqrt{x}+1}\ge0\) ; \(\forall x\)

\(\Rightarrow C\ge-8\)

\(C_{min}=-8\) khi \(x=0\)

2^x : 16 = 2²⁰¹⁶

2^x : 2⁴ = 2²⁰¹⁶

2^x = 2²⁰¹⁶ . 2⁴ = 2²⁰²⁰

Vậy x = 2020

2x : 24 = 22016

2x = 22016 . 24 

= 22020 (2020)

\(2^x:16=2^{2016}\)

\(2^x:2^4=2^{2016}\)

\(x:4=2016\)

\(x=2016\cdot4\)

\(x=8064\)

\(f'\left(x\right)=2-\dfrac{\pi}{2}sin\left(\dfrac{\pi x}{3}\right)=\dfrac{1}{2}\left(4-\pi sin\left(\dfrac{\pi x}{2}\right)\right)\)

Do \(\left|\pi sin\left(\dfrac{\pi x}{2}\right)\right|\le\pi< 4\Rightarrow f'\left(x\right)>0\) ; \(\forall x\)

\(\Rightarrow f\left(x\right)\) đồng biến trên R

\(\Rightarrow f\left(x\right)_{min}+f\left(x\right)_{max}=f\left(-2\right)+f\left(2\right)=-4+cos\left(-\pi\right)+4+cos\left(\pi\right)=-2\)

Ta có  2 + 1 2017 = C 2017 0 .2 2017 + C 2017 1 .2 2016 + ... + C 2017 2017 .2 0

2 − 1 2017 = C 2017 0 .2 2017 + C 2017 1 .2 2016 . − 1 + ... + C 2017 2017 .2 0 . − 1 2017

Trừ từng vế hai đẳng thức trên ta được:

3 2017 − 1 = 2 C 2017 1 .2 2016 + C 2017 3 .2 2014 + ... + C 2017 2017 .2 0

Vậy  M = 3 2017 − 1 2

Chọn đáp án D.

\(A=x^2+5y^2+4xy+2x+12\)

\(A=x^2+4xy+4y^2+2\left(x+2y\right)+1+y^2-4y+4+7\)

\(A=\left(x+2y\right)^2+2\left(x+2y\right)+1+\left(y-2\right)^2+7\)

\(A=\left(x+2y+1\right)^2+\left(y-2\right)^2+7\)

Vì \(\left(x+2y+1\right)^2\ge0\forall x;y\)và \(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=2\end{cases}}}\)

Vậy....