a) \(5^{300}\) và \(3^{500}\)

\(\Rightarrow5^{300}< 3^{555}\)

b) \(2^{400}\) và \(4^{200}\)

\(\Rightarrow2^{400}=4^{200}\)

c) \(5^{300}\) và \(3^{453}\)

\(\Rightarrow5^{300}>3^{453}\)

B

B

B

mk ko thấy đề

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

Cô làm đ

Bài 1:

a) \(\dfrac{-17}{36}\) và \(\dfrac{23}{-48}\) 

\(\dfrac{-17}{36}=\dfrac{-17.4}{36.4}=\dfrac{-68}{144}\) 

\(\dfrac{23}{-48}=\dfrac{-23}{48}=\dfrac{-23.3}{144.3}=\dfrac{-69}{144}\) 

Vì \(\dfrac{-68}{144}>\dfrac{-69}{144}\) nên \(\dfrac{-17}{36}>\dfrac{23}{-48}\) 

b) \(\dfrac{-1}{3}\) và \(\dfrac{2}{5}\) 

Vì \(\dfrac{-1}{3}\) là số âm mà \(\dfrac{2}{5}\) là số dương nên \(\dfrac{-1}{3}< \dfrac{2}{5}\) 

c) \(\dfrac{2}{7}\) và \(\dfrac{5}{4}\) 

Vì \(\dfrac{2}{7}< 1\) mà \(\dfrac{5}{4}>1\) nên \(\dfrac{2}{7}< \dfrac{5}{4}\) 

d) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\) 

\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\) 

\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\) 

Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

Bài 2:

\(\dfrac{5}{2}-\left(1\dfrac{3}{7}-0,4\right)=\dfrac{5}{2}-\dfrac{10}{7}-\dfrac{2}{5}=\dfrac{47}{70}\) 

a) Ta có: 

\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)

\(A=-21\cdot26\)

\(A=-546\)

\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)

\(B=2\cdot12\cdot5\)

\(B=2\cdot60\)

\(B=120\)

Mà: \(120>-546\)

\(\Rightarrow B>A\)

1. 

a)=1/3-[(-5/4)-5/8]

=1/3-(-15/8)=53/24

b)=5/9:(-3/22)+5/9:(-3/5)

=5/9*22/-3+5/9*5/-3=-110/27+-25/27=5

2

a)Ta có 3³⁹<3⁴⁰=9²⁰<11²⁰<11²¹

 nên 3³⁹<11²¹

^{b)Ta có} /3,4-x/ lớn hơn hoặc bằng 0 Với mọi x thuộc R

          => -/3,4-x/ bé hơn hoặc bằng 0 Với mọi x thuộc R

           => 0,5-/3,4-x/ bé hơn hoặc bằng 0,5 Với mọi x thuộc R

  Dấu = xảy ra khi 3,4-x=0

                        =>x=3,4

 Vậy GTLN của A = 0,5 khi x=3,4

a: \(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{9}{3}\)

\(\dfrac{7}{3}+\dfrac{2}{3}=\dfrac{9}{3}\)

=>\(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{7}{3}+\dfrac{2}{3}\)

\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}\)

=>\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{4}{5}+\dfrac{3}{5}\)

b: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{7+16}{9}=\dfrac{23}{9}\)

\(\dfrac{16}{9}+\dfrac{7}{9}=\dfrac{16+7}{9}=\dfrac{23}{9}\)

Do đó: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{16}{9}+\dfrac{7}{9}\)