Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(A=\frac{5^{2020}+1}{5^{2020}+1}=1\)
\(B=\frac{5^{2019}+1}{5^{2020}+1}< 1\)
=> B < A
(125 - 2x) . 6 = 126 - 12
(125 - 2x) . 6 = 114
(125 - 2x) = 114 : 6
(125 - 2x) = 19
2x = 125 - 19
2x = 106
x = 3
(54 : 52 - 4x) .2 - 13 = 44 : 42 + 5
(54 : 52 - 4x) .2 - 13 = 21
(54 : 52 - 4x) . 2 = 21 + 13
(54 : 52 - 4x) . 2 = 34
(54 : 52 - 4x) = 34 : 2
(54 : 52 - 4x) = 17
25 - 4x = 17
4x = 8
x = 2
a) 5x+x+1=\(\dfrac{125}{25}\)
\(\leftrightarrow\) 52x+1 =51
\(\leftrightarrow\) 2x+1=1
\(\leftrightarrow\)2x=0
\(\leftrightarrow\) x=0
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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a: =5-78*32
=5-2496
=-2491
b: \(=6\left(9-6\right)=6\cdot3=18\)
c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)
d: \(=181+3-84+8\cdot25\)
=100+200
=300
e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)
=14000-1
=13999
f: \(=3^3+25\cdot8-1=26+200=226\)
g: \(=3+2^4+1=16+4=20\)
h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)
Ta có: \(5^{4x}:5^5=5^{2020}:5^{2019}\)
\(\Leftrightarrow5^{4x-5}=5^1\)
\(\Leftrightarrow4x-5=1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\left(L\right)\)
Vậy \(x=\varnothing\)