Tính: M = \(2^{2017}-\left(2^{2016}+2^{2015}+...+2^1+2^0\right)\)
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28 tháng 7 2018
\(A=2^{2017}-(2^{2016}+2^{2015}+......+2^1+2^0)\)
Đặt \(B=2^{2016}+2^{2015}+.....+2^1+2^0\)
\(\Rightarrow2B=2^{2017}+2^{2016}+....+2^1+2^0\)
\(\Rightarrow2B-B=(2^{2017}+2^{2016}+...+2^0)-(2^{2016}+2^{2015}+...+2^1+2^0)\)
\(\Rightarrow B=2^{2017}-2^0\)
\(\Rightarrow A=2^{2017}-(2^{2017}-1)\)
\(\Rightarrow A=1\)
28 tháng 7 2018
2A = 22018 - (22017 + 22016 + ....+ 21)
2A - A = [22018 - (22017 + 22016 + ....+ 21 )] - [22017 - (22016 + 22015 +..... + 21 + 20)
A = 22018 - 22017 - 22017 - 1
A = 22018 - (22017 +22017 +1)
A = 22018 - (22018 +1 )
A = -1
12 tháng 11 2017
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
BT
14 tháng 2 2017
Ta có: 1+(1+2)+(1+2+3)+...+(1+2+3+...+2017)=2017x1+2016x2+2015x3+...+2x2016+1x2017
=> K-2016=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017x1+2016x2+2015x3+...+2x2016+1x2017}\)=\(\frac{2017x1+2016x2+2015x3+...+2x2016+1x2017}{2017x1+2016x2+2015x3+...+2x2016+1x2017}=1\)
=> K=2016+1=2017
14 tháng 2 2017
Toán tiếng anh hả bạn
Bài này thì bạn mình có thể giải được
Thank you
1 tháng 5 2018
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
HP
14 tháng 5 2016
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
14 tháng 5 2016
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
2 tháng 3 2016
Tử số bằng mẫu số
K-2016=1
K=2017
Muốn biết tại sao tử= mẫu thì tích nha
2 tháng 3 2016
\(K-2016=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)
\(K-2016=\frac{1\times2017+2\times2016+3\times2015+...+2017\times1}{2017\times1+2016\times2+2015\times3+...+2017\times1}\)
\(K-2016=1\)
\(\Rightarrow K=1+2016\)
\(\Rightarrow K=2017\)
28 tháng 3 2020
\(\Leftrightarrow4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy M=1
Ta có :
\(M=2^{2017}-\left(2^{2016}+2^{2017}+...............+2+1\right)\)
Đặt :
\(A=2^{2016}+2^{2015}+................+2+1\)
\(\Leftrightarrow2A=2^{2017}+2^{2016}+2^{2015}+............+2^2+2\)
\(\Leftrightarrow2A-A=\left(2^{2017}+2^{2016}+........+2\right)-\left(2^{2016}+2^{2015}+..........+1\right)\)
\(\Leftrightarrow A=2^{2017}-1\)
\(\Leftrightarrow M=2^{2017}-A\)
\(\Leftrightarrow M=2^{2017}-\left(2^{2017}-1\right)\)
\(\Leftrightarrow M=2^{2017}-2^{2017}+1\)
\(\Leftrightarrow M=0+1=1\)
\(M=2^{2017}-\left(2^{2016}+2^{2015}+...+2^1+2^0\right)\)
Đặt :
\(S=2^{2016}+2^{2015}+...+2^1+2^0\)
\(\Rightarrow S=2^0+2^1+...+2^{2015}+2^{2016}\)
\(\Rightarrow2S=2\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)
\(\Rightarrow2S=2^1+2^2+...+2^{2016}+2^{2017}\)
\(\Rightarrow2S-S=\left(2^1+2^2+...+2^{2016}+2^{2017}\right)-\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)
\(\Rightarrow S=2^{2017}-1\)
Thay S vào M ta có:
\(M=2^{2017}-\left(2^{2017}-1\right)\)
\(M=2^{2017}-2^{2017}+1=1\)