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\(-\frac{1}{13}+\frac{15}{21}+\frac{\left(-12\right)}{13}+\frac{2}{7}+\frac{2020}{2021}\)
\(=-\frac{1}{13}+\frac{15}{21}-\frac{12}{13}+\frac{6}{21}+\frac{2020}{2021}\)
\(=\left(-\frac{1}{13}-\frac{12}{13}\right)+\left(\frac{15}{21}+\frac{6}{21}\right)+\frac{2020}{2021}\)
\(=-1+1+\frac{2020}{2021}=\frac{2020}{2021}\)
Bài 1: Bạn xem lại đã viết đúng đề chưa vậy.
Bài 2:
$P=29-|16+3.2|+1=29-|22|+1=29-22+1=7+1=8$
Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
a) \(2021^{2020}-2021^{2019}=2021^{2019}.\left(2021-1\right)=2021^{2019}.2020\)
b) Ta có :\(7x-140=3.7^2\)
\(\implies\) \(7x-140=3.49\)
\(\implies\) \(7x-140=147\)
\(\implies\) \(7x=287\)
\(\implies\) \(x=41\)
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: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
thực hiện phép tính sau một cách hợp lí:
C=(1+2+3)+(3+4+5)+(5+6+7)+..........+(97+98+99)+(99+100+101)
Ko ghi đề
C = (3.2)+(3.4)+(3.6)+...+(3.98)+(3.100)
C = 3 . (2+4+6+...+100)
C = 3 . 2550
C = 7650
Ko theo công thức nha @@
Nhớ k mk nha ^^
\(\left(\frac{1}{2}\right)^5.2^5-\left(\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2021}\right)\)
\(=\frac{1^5}{2^5}.2^5-\left(\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2021}\right)\)
\(=\frac{1^5.2^5}{2^5}-\left(\frac{2020.2021}{2019.2020.2021}-\frac{2019.2021}{2019.2020.2021}+\frac{2019.2020}{2019.2020.2021}\right)\)
\(=1^5-\left(\frac{2020.2021-2019.2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{\left(2020-2019\right).2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1.2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020.\left(1+2019\right)}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020.2020}{2019.2020.2021}\right)\)
\(=1-\frac{1+2020}{2019.2021}\)
\(=1-\frac{2021}{2019.2021}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
Chúc bạn học tốt
\(2\cdot\left[\left(7-3^{2021}:3^{2020}\right):2^2+99\right]-100\)
\(=2\cdot\left[\left(7-3\right):4+99\right]-100\)
\(=2\cdot\left(4:4+99\right)-100\)
\(=2\cdot\left(1+99\right)-100\)
\(=2\cdot100-100\)
\(=200-100\)
\(=100\)
#Urushi☕