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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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
Ta có: ( Sửa đề )
\(A=4+4^2+4^3+...+4^{2021}+4^{2022}\)
\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)
\(A=20+4^2.\left(4+4^2\right)+...+4^{2020}.\left(4+4^2\right)\)
\(A=20+4^2.20+...+4^{2020}.20\)
\(A=20.\left(1+4^2+...+4^{2020}\right)\)
Vì \(20⋮20\) nên \(20.\left(1+4^2+...+4^{2020}\right)\)
Vậy \(A⋮20\)
\(#WendyDang\)
a) 38 : 34 + 22 x 23 = 113
b) 3 x 42 - 2 . 3 = 42
c) 46 . 34 . 95 : 612 = 9
d) 212 . 14 . 125 : 353 . 6 = 108
e) 453 . 204 .182 : 1805 = 25
#BạcHà#
bn ơi ghi cả cách làm nữa bn
ai lm lun cách làm mk sẽ k cho 2 k lun
a,( 393+390) : (317. 373)
= (33+1). 390 : 390
= 33+1
=27+1
=28
b,(556+57) : (549+1)
=57. (549+1) : (549+1)
=57= 78125
c,(722+721+720) ; (25+24+32)
= 720. (72+71+1) : [24. (2+1)+32 ]
= 720. 57 : [ 24. 3 +32 ]
= 720. 57 : ( 24+3) . 3
= 720. 57 : 19 . 3
= 720. 57 : 57
= 720
Bài 1:
a) 02002 < 02023
b) 20220 = 20230
c) 549 < 5510
d) ( 4 + 5 )3 > 42 + 52
đ) 92 - 32 > ( 9 - 3 )2
Bài 2:
a) 32 x 43 - 32 + 333
= 9 x 64 - 9 + 333
= 576 - 9 + 333
= 567 + 333
= 900
b) 5 x 43 + 24 x 5 + 410
= 5 x 64 + 24 x 5 + 1
= 5 x ( 64 + 24 ) + 1
= 5 x 88 + 1
= 440 + 1
= 441
c) 23 x 42 + 32 x 5 - 40 x 12023
= 8 x 16 + 9 x 5 - 40 x 1
= 128 + 45 - 40
= 133
Bài 1 :
a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)
b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)
c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)
d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)
đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)
Đặt A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
3A=3/2.5+3/5.8+....+3/17.20
3A=1/2-1/5+1/5-1/8+...+1/17-1/20
3A=1/2-1/20
3A=9/20
2)
Giữ nguyên p/s 1/2^2
Ta có:1/3^2<1/2.3
1/4^2<1/3.4
...............
1/n^2<1/(n-1).n
=>1/3^2+1/4^2+...+1/n^2<1/2.3+1/3.4+...+1/(n-1).n
=>1/3^2+1/4^2+.....+1/n^2<1/2-1/3+1/3-1/4+.........+1/n-1-1/n
=>1/2^2+1/3^2+.....+1/n^2<1/2^2+1/2-1/n
=>1/2^2+1/3^2+....+1/n^2<3/4-1/n<3/4
3)
2B=2/3.5+2/5.7+....+2/47.49+2/49.51
2B=1/3-1/5+1/5-1/7+.....+1/47-1/49+1/49-1/51
2B=1/3-1/51
2B=16/51
B=16/51:2
B=8/51
A=1+1/2+1/2^2+...+1/2^2010
2A=2+1+1/2+....+1/2^2009
2A-A=(2+1+1/2+...+1/2^2009)-(1+1/2+1/2^2+....+1/2^2010)
A=2-1/2^2010
Ta có :
\(a=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow2a=8+2^3+2^4+2^5+...+2^{20}+2^{21}\)
\(\Rightarrow2a-a=a=4+2^{21}-2^2\)
\(\Rightarrow a=4+2^{21}-4\)
\(\Rightarrow a=2^{21}\)