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đề có pải là A=\(\frac{19^{30}+5}{19^{31}+5}\) ; B=\(\frac{19^{31}+5}{19^{32}+5}\) PẢI KO BẠN
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 \(30^{31}+5< 30^{32}+5\Rightarrow\frac{90}{30^{31}+5}>\frac{90}{30^{32}+5}\)
\(\Rightarrow1+\frac{90}{30^{31}+5}>1+\frac{90}{30^{32}+5}\)
\(\Rightarrow19A>19B\Rightarrow A>B\)
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) \(81^{40}=\left(3^4\right)^{40}=3^{160}\)
\(27^{14}=\left(3^3\right)^{14}=3^{42}\)
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 ^^
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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)