so sánh mà 0 thực hiện phép tính
2021^0 và 1^2021
21^15 và 27^5 .49^8
3^2n và 2^3n
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a) 17 + 15 + (-27)
= (17 - 27) + 15
= -10 + 15
= 5
b) 13.75 + 25.13 - 300
= 13.(75 + 26) - 300
= 13.100 - 300
= 1300 - 300
= 1000
c) 80 - (4.5² - 5.2³) + 2021⁰
= 80 - (4.25 - 5.8) + 1
= 80 - (100 - 40) + 1
= 80 - 60 + 1
= 20 + 1
= 21
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
32n và 23n
Ta có : 32n = (32)n = 9n
23n = (23)n = 8n
9n > 8n => 32n > 23n
Ta có :
\(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n\)
Vì \(9>8\)\(\Rightarrow\)\(9^n>8^n\)
Hay \(3^{2n}>2^{3n}\)
Vậy \(3^{2n}>2^{3n}\)
_Chúc bạn học tốt_
\(-\frac{1}{13}+\frac{15}{21}+\frac{\left(-12\right)}{13}+\frac{2}{7}+\frac{2020}{2021}\)
\(=-\frac{1}{13}+\frac{15}{21}-\frac{12}{13}+\frac{6}{21}+\frac{2020}{2021}\)
\(=\left(-\frac{1}{13}-\frac{12}{13}\right)+\left(\frac{15}{21}+\frac{6}{21}\right)+\frac{2020}{2021}\)
\(=-1+1+\frac{2020}{2021}=\frac{2020}{2021}\)
c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
\(\dfrac{a}{b}=\dfrac{a\left(b+2021\right)}{b\left(b+2021\right)}=\dfrac{ab+2021a}{b\left(b+2021\right)}\\ \dfrac{a+2021}{b+2021}=\dfrac{ab+2021b}{b\left(b+2021\right)}\)
Vì \(b>0\Rightarrow b\left(b+2021\right)>0\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2021}{b+2021}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2021}{b+2021}=1\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)
a) Ta có : 20210 = 1 ; 12021 = 1
Vậy 1 = 1 nên 20210 = 12021
b) Ta có : 2115 = ( 3 . 15 )15 = 315 . 1515
275 . 498 = ( 33 )5 . ( 72 )8 = 315 . 148
Vì 1515 > 148 nên 315 . 1515 > 315 . 148 hay 2115 > 275 . 498
c) Ta có : 32n = ( 32 )n = 9n
23n = ( 23 )n = 8n
Vì 8 < 9 nên 9n > 8n hay 32n > 23n