Thực hiện phép tính:
22×3-(12012+20120)÷(-2)
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\(a,=4\cdot3-\left(1+1\right):2=12-2:2=12-1=11\\ b,=\left[47-\left(736:2^4\right)\right]\cdot2013=\left(47-736:16\right)\cdot2013=\left(47-46\right)\cdot2013=2013\)
f: =-46-64-91+220
=220-110-91
=110-91=19
g: =4-2:2=3
h: ={47-736:2^4}*2013
=2013(47-184)
=-137*2013=-275781
a: \(\left[600-\left(40:2^3+3\cdot5^3\right)\right]:5\)
\(=\left[600-5-375\right]:5\)
\(=44\)
b: \(16\cdot12^2-\left(4\cdot23^2-59\cdot4\right)\)
\(=16\cdot144-4\cdot\left(23^2-59\right)\)
\(=2304-4\cdot470\)
\(=424\)
c: Ta có: \(2^{100}-\left(1+2+2^2+2^3+...+2^{99}\right)\)
\(=2^{100}-2^{100}+1\)
=1
d: Ta có: \(169\cdot2011^0-17\cdot\left(83-1702:23+1^{2012}\right)+2^7:2^4\)
\(=169-17\cdot\left(83-74+1\right)+2^3\)
\(=177-17\cdot10\)
=7
\(2011-5.5^2-\left(7^3:7+2012^0\right)\\ =2011-5^3-\left(7^2+1\right)\\ =2011-125-50=1836\)
\(120:520:\left(500-5^3+35.7\right)\\ =120:520:\left(500-125+245\right)\\ =120:520:620=\dfrac{3}{8060}\)
\(2011-5\times5^2-\left[7^3:7+2012^0\right]\\ =2011-5^3-\left(7^2+1\right)\\ =2011-125-50\\ =1836\\ 120:520:\left[500-5^3+35\times7\right]\\ =120:520:\left[500-125+35\times7\right]\\ =120:520:620\\ =\dfrac{3}{8060}\)
Lời giải:
$A=2^{50}-2^{49}-...2^2-2$
$-A=2+2^2+2^3+...+2^{49}-2^{50}$
$-2A=2^2+2^3+2^4+...+2^{50}-2^{51}$
$-A-(-2A)=(2-2^{50})-(2^{50}-2^{51})$
$A=2-2^{50}-2^{50}+2^{51}=2-2^{51}+2^{51}=2$
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
22×3-(12012+20120)÷(-2)
=4×3-(1+1)÷(-2)
=4×1÷(-2)
=4÷(-2)=-2
=4x3-2:(-2)
=12-(-1)
=13