giúp mk với mọi người ơi

Đặt 19³¹ = a ( cho đơn giản nha)

\(A=\frac{\frac{a}{19}+5}{a+5}=\frac{a+95}{19\left(a+5\right)}\)

\(B=\frac{a+5}{19a+5}\)

Ta có

\(B-A=\frac{a+5}{19a+5}-\frac{a+95}{19\left(a+5\right)}=-\frac{1620a}{19\left(a+5\right)\left(19a+5\right)}< 0\)

Vậy A > B

Cách khá nhé

Ta có

\(19A=\frac{30^{31}+19.5}{30^{31}+5}=1+\frac{90}{30^{31}+5}\)

\(19B=\frac{30^{32}+19.5}{30^{32}+5}=1+\frac{90}{30^{32}+5}\)

Vì \(30^{31}+5< 30^{32}+5\Rightarrow\frac{90}{30^{31}+5}>\frac{90}{30^{32}+5}\)

\(\Rightarrow1+\frac{90}{30^{31}+5}>1+\frac{90}{30^{32}+5}\)

\(\Rightarrow19A>19B\Rightarrow A>B\)

mấy bạn trên làm đúng rồi.

Ta có : \(A=\frac{19^{30}+15}{19^{31}+15}\)

\(\Rightarrow19A=\frac{19^{31}+285}{19^{31}+15}=\frac{19^{31}+15+270}{19^{31}+15}=1+\frac{270}{19^{31}+15}\)

Lại có \(B=\frac{19^{31}+15}{19^{32}+15}\)

\(\Rightarrow19B=\frac{19^{32}+285}{19^{32}+15}=\frac{19^{32}+15+270}{19^{32}+15}=1+\frac{270}{19^{32}+15}\)

Vì \(\frac{270}{19^{32}+15}< \frac{270}{19^{31}+15}\Rightarrow1+\frac{270}{19^{32}+5}< 1+\frac{270}{19^{31}+15}\Rightarrow19B< 19A\Rightarrow B< A\)

\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)

\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)

từ (1) and (2)

=>19A>19B

=>A>B

a) \(81^{40}=\left(3^4\right)^{40}=3^{160}\)

   \(27^{14}=\left(3^3\right)^{14}=3^{42}\)

Vì \(3^{160}>3^{42}\) => \(81^{40}>27^{14}\)

b) \(5^{64}=5^{4.16}=625^{16}\)

   \(3^{96}=3^{6.16}=729^{16}\)

Vì \(625^{16}< 729^{16}\)=> \(5^{64}< 3^{96}\)

c) \(125^{12}=\left(5^3\right)^{12}=5^{36}\)

   \(25^{10}=\left(5^2\right)^{10}=5^{20}\)

Vì \(5^{36}>5^{20}\)=> \(125^{12}>25^{10}\)

T_i_c_k nha,mơn bạn nhìu ^^

a, 81⁴⁰>27¹⁴

b, 5⁶⁴<3⁹⁶

c, 125¹²<25¹⁰

a. 81⁴⁰  lớn hơn 27¹⁴

b. 5⁶⁴ bé hơn 3⁹⁶

^c. 125¹² lớn hơn 25¹⁰

k cho mk nhá

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