so sánh
2011.2013+2012.2014
\(2012^2+2013^2-2\)
Những câu hỏi liên quan
so sánh 2011.2013+2012.2014 và 2012+2013^2-2
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
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Thực hiện tính :
a) A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/2013(1+2+3+..+2013)
b) B = 1-3/7.3+2-4/2.4+3-5/3.5+4-6/4.6+....+2011-2013/2011.2013+2012-2014/2012.2014-2013+2014/2013.2014
tính
A=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/2013(1+2+3+4+...+2013)
B=(1-3)/(1.3)+(2-4)/(2.4)+(3-5)/(3.5)+(4-6)/(4.6)+...+(2011-2013)/(2011.2013)+(2012-2014)/(2012.2014)-(2013+2014)/(2013.2014)
thứ 7 mình nộp ai làm nhanh mình tích cho
nhớ giải chi tiết
So sánh
a, 2011.2013+2012.2014 và 2012^2+2013^2-2
b, (9-1)(9^2+1)(9^4+1)(9^8+1)(9^16+1)(9^32+1) và 9^64-1
c, x-y/x+y và x^2-y^2/x^2+xy+y^2 với x>y>0
Giúp mình nha!!!
a) \(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{64}-1\right)\)
\(=\dfrac{9^{64}-1}{10}\)
Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)
Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)
\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)
c) Ta có:
\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)
Vì x>y>0, ta có:
\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)
Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)
Từ (1), (2) và (3) suy ra:
\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)
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a) Ta có:
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
Vậy 2011.2013+2012.2014 = 20122 + 20132 - 2
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Thực hiện phép tính :
A = \(\frac{1-3}{1.3}+\frac{2-4}{2.4}+\frac{3-5}{3.5}+..........+\frac{2012-2014}{2012.2014}-\frac{2013+2014}{2013.2014}\)
giúp mình nha
Thực hiện phép tính
a) A= \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)\)\(+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)
b) B=\(\dfrac{1-3}{1.3}+\dfrac{2-4}{2.4}+\dfrac{3-5}{3.5}+\dfrac{4-6}{4.6}+...+\dfrac{2011-2013}{2011.2013}+\dfrac{2012-2014}{2012.2014}-\dfrac{2013+2014}{2013.2014}\)
So sánh hai số A và B mà không tính kết quả cụ thể A=20132 và B=2012.2014
A= 2013. ( 2021 + 1 ) = 2013 . 2012 + 2013
B = 2012 . 2014 = 2012 . ( 2013 + 1 ) = 2012 . 2013 + 2012
Vì 2013 > 2012 ==> A > B
So sánh phân thức A=\(\frac{2013^2-2012^2}{2013^2+2012^2}\) với B=\(\frac{2013-2012}{2013+2012}\)
tính hợp lí:
20134 -.2012.2014.(20132+1)
\(=2013^4-\left(2012\cdot2014\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^2-1\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^4-1\right)\\ =1\)
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