a: \(0,5^{1000}=\left(0,5^5\right)^{200}=0,03125^{200}\)

mà \(0,03125< 0,625\)

nên \(0,5^{1000}< 0,625^{200}\)

c: \(A=2+2^2+...+2^{2022}\)

=>\(2A=2^2+2^3+...+2^{2023}\)

=>\(2A-A=2^2+2^3+...+2^{2023}-2-2^2-...-2^{2022}\)

=>\(A=2^{2023}-2\)

=>A<B

e: \(2020A=\dfrac{2020^{2024}-2020}{2020^{2024}-1}=1-\dfrac{2019}{2020^{2024}-1}\)

\(2020B=\dfrac{2020^{2024}+2020}{2020^{2024}+1}=1+\dfrac{2019}{2020^{2024}+1}\)

Vì \(-\dfrac{2019}{2020^{2024}-1}< 0< \dfrac{2019}{2020^{2024}+1}\)

nên \(-\dfrac{2019}{2020^{2024}-1}+1< \dfrac{2019}{2020^{2024}+1}+1\)

=>2020A<2020B

=>A<B

d: \(\left(-\dfrac{3}{2}\right)^{2024}=\left(\dfrac{3}{2}\right)^{2024};\left(-2\right)^{2024}=2^{2024}\)

mà 3/2<2

nên \(\left(-\dfrac{3}{2}\right)^{2024}< 2^{2024}\)

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$\left(2\frac13-1,5\right):\left(-6\frac16+5\frac12\right)+2,75$

$=\left(\frac73-\frac32\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}$

$=\frac56:\frac{-2}{3}+\frac{11}{4}$

$=-\frac54+\frac{11}{4}=\frac32$

\(a.\dfrac{3}{7}=\dfrac{2x+1}{3x+5}\\ 3\left(3x+5\right)=7\left(2x+1\right)\\ 9x+15=14x+7\\ 14x-9x=15-7\\ 5x=8\\ x=\dfrac{8}{5}\\ b.\dfrac{x+1}{x-2}=\dfrac{3}{4}\\ 3\left(x-2\right)=4\left(x+1\right)\\ 3x-6=4x+4\\ 4x-3x=-6-4\\ x=-10\\ c.\dfrac{2x+3}{7}=\dfrac{4x-1}{15}\\ 15\left(2x+3\right)=7\left(4x+1\right)\\ 30x+45=28x+7\\ 30x-28x=7-45\\ 2x=-38\\ x=\dfrac{-38}{2}=-19\\ d.\dfrac{6x-5}{-7}=\dfrac{5x-3}{-5}\\ -5\left(6x-5\right)=-7\left(5x-3\right)\\ -30x+25=-35x+21\\ -30x+35x=21-25\\ 5x=-4\\ x=-\dfrac{4}{5}\)

\(a,\dfrac{3}{7}=\dfrac{2x+1}{3x+5}\\\Rightarrow3x+15=14x+7\\ \Rightarrow3x-14x=7-15 \\ \Rightarrow-11x=-8\\ \Rightarrow x=\dfrac{8}{11}.\\ b,\dfrac{x+1}{x-2}=\dfrac{3}{4}\\ \Rightarrow4x+4=3x-6\\ \Rightarrow4x-3x=-6-4\\ \Rightarrow x=-10.\\ c,\dfrac{2x+3}{7}=\dfrac{4x-1}{15}\\ \Rightarrow30x+45=28x-7\\\Rightarrow30x-28x=-7-45\\ \Rightarrow2x=-52\\ \Rightarrow x=-26.\)

\(d,\dfrac{6x-5}{-7}=\dfrac{5x-3}{-5}\\ \Rightarrow-30x+25=-35x+21\\ \Rightarrow-30x+35x=21-25\\ \Rightarrow5x=-4\\ \Rightarrow x=-\dfrac{4}{5}.\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) 

a) \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{dk}{dk+d}=\dfrac{c}{c+d}\) 

b) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{2c+5d}{3c-4d}\)

c) \(\dfrac{2018a-2019b}{2019c+2020d}=\dfrac{2018bk-2019b}{2019dk+2020d}=\dfrac{b\left(2018k-2019\right)}{d\left(2019k+2020\right)}=\dfrac{b}{d}\cdot\dfrac{2018k-2019}{2019k+2020}\) (1) 

Mà: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{b}{d}=\dfrac{c}{a}\) 

\(\left(1\right)=\dfrac{c}{a}\cdot\dfrac{2018k-2019}{2019k+2020}=\dfrac{2018ck-2019c}{2019ak+2020a}=\dfrac{2018ck-2019dk}{2019ak+2020bk}\\ =\dfrac{k\left(2018c-2019d\right)}{k\left(2019a+2020b\right)}=\dfrac{2018c-2019d}{2019a+2020b}\)

(5.3^5) + 17.3^4) : 6^2

= (5.243 + 17.81) : 36

= (1215 + 1377) : 36

= 2592 : 36

= 72

\(x^2-64\cdot0,5\cdot x+3=0\\ \Leftrightarrow x^2-32x+256=253\\ \Leftrightarrow x^2-32x+16^2=253\\ \Leftrightarrow\left(x-16\right)^2=253\\ \Leftrightarrow\left[{}\begin{matrix}x-16=\sqrt{253}\\x-16=-\sqrt{253}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{253}+16\\x=-\sqrt{253}+16\end{matrix}\right.\)

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