TL:

C=\(\frac{2020}{-\left(x^2+2x-2020\right)}\) 

 =\(\frac{2020}{-\left(x^2+2x+1-2021\right)}=\frac{2020}{-\left(x+1\right)^2+2021}\) 

Để Cmin thì \(-\left(x+1\right)^2+2021\) lớn nhất

vì \(-\left(x+1\right)^2+2021\le2021\) =>-(x+1)+2021 lớn nhất =2021

vậy C_min=\(\frac{2020}{2021}\)

mik đang cần gấp

Lời giải:

$2^x+2^{x+1}+2^{x+2}+....+2^{x+2020}=2^{x+2024}-8$

$2^x(1+2+2^2+...+2^{2020})=2^{x+2024}-8$

$2^x(2+2^2+2^3+...+2^{2021})=2^{x+2025}-16$

$\Rightarrow 2^x(2+2^2+2^3+...+2^{2021})- (2^x(1+2+2^2+...+2^{2020}))=2^{x+2025}-16-(2^{x+2024}-8)$

$\Rightarrow 2^x(2^{2021}-1)=2^{x+2025}-2^{x+2024}-8$

$\Rightarrow 2^x(2^{2021}-1)=2^{x+2024}(2-1)-8$

$\Rightarrow 2^{x+2021}-2^x=2^{3+2021}-2^3$

$\Rightarrow x=3$

b: \(\Leftrightarrow\left\{{}\begin{matrix}x-7y=0\\11x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{11}\\y=\dfrac{x}{7}=\dfrac{5}{77}\end{matrix}\right.\)

Lời giải:
a. Bạn cần viết đề bằng công thức toán để đề được rõ ràng hơn.

b. Ta có:

$(7y-x)^{2020}\geq 0$ với mọi $x,y$

$|5-11x|^{2021}\geq 0$ với mọi $x,y$

Do đó để tổng của chúng bằng $0$ thì:

$(7y-x)^{2020}=|5-11x|^{2021}=0$

$\Leftrightarrow x=\frac{5}{11}; y=\frac{5}{77}$

Lời giải:

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2020}=2^{2024}-8$

$2^x(1+2+2^2+...+2^{2020})=2^{2024}-8(1)$

$2^x(2+2^2+2^3+...+2^{2021})=2^{2025}-16(2)$

Lấy $(2)$ trừ $(1)$ ta có:

$2^x(2^{2021}-1)=2^{2025}-16-(2^{2024}-8)=2^{2024}(2-1)-8$

$2^x(2^{2021}-1)=2^{2024}-8=2^3(2^{2021}-1)$

$\Rightarrow 2^x=2^3$
$\Rightarrow x=3$
 

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)