A=\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\dfrac{1-3-5-7-...-49}{89}\)
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\dfrac{1-3-5-7-...-49}{89}\)
thực hiện phép tính
Đặt \(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}\)
Đặt \(B=1-3-5-7-..-49\)
\(=1-\left(3+5+7+...+49\right)\)
\(=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}\)
\(=1-624\)
\(=-623\)
\(\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}\)
Vậy: \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}\)
Xét \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}.\dfrac{45}{196}\)
=\(\dfrac{9}{196}\)
Xét \(\dfrac{1-3-5-7-..-49}{89}\)
=\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)
CT tính sl số hạng (số cuối - số đầu ):2+1
số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24
Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2
=> tổng = [(3+49).24]:2=624
=>\(\dfrac{1-624}{89}\)
=\(\dfrac{-623}{89}\)
=-7
từ đó ta có \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)
Tính :
c) C = \(\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
d) D = \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
Tính:
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
Giải chi tiết dùm mik nha. Thankss
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)
a) Tìm \(n\in N\)* :
+) \(\dfrac{1}{8}.16^n=2^n\)
+) \(27< 3^n< 243\)
b) Tính : \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
Bài 1 :
a) +) \(\dfrac{1}{8}\cdot16^n=2^n\)
\(\Leftrightarrow\dfrac{1}{8}=\dfrac{2^n}{16^n}\)
\(\Rightarrow\dfrac{1}{8}=\dfrac{1}{8}^n\)
Vậy n = 1.
+) \(27< 3^n< 243\)
\(\Leftrightarrow3^3< 3^n< 3^5\)
Vậy n = 4.
Bài 2 : \(\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}=-\dfrac{45}{28}\)
\(\left(\dfrac{1}{4}.9+\dfrac{1}{9.14}+\dfrac{1}{14.19}+....+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
GIÚP với các bn!
=\(\dfrac{1}{5}\).(\(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+....+\dfrac{5}{44.49}\)).\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)
=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\).\(\dfrac{1-624}{89}\)
=\(\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\).(-7)
=\(\dfrac{1}{5}\).\(\dfrac{45}{196}\).(-7)=\(\dfrac{-9}{28}\)
A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)
C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)
E=13+23+...+103=3025
Tính F=23+42+63+...+203=?
Thực hiện phép tính hợp lí nếu có thể:
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
\(=\) \(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\)
\(=\) \(\dfrac{1}{4}-\dfrac{1}{49}\)
\(=\) \(\dfrac{49}{196}-\dfrac{4}{196}\)
\(=\) \(\dfrac{45}{196}\)
Ta có : \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
= 5.\(\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
= 5. \(\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
=5. \(\dfrac{45}{196}\)
=\(\dfrac{225}{196}\)
\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{44.49}=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+...+\dfrac{5}{44.49}\right)=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)=\dfrac{1}{5}.\dfrac{45}{196}=\dfrac{9}{196}\)
1. Tìm giá trị n nguyên dương :
\(\dfrac{1}{8} . 16^n=2^n\)
2. Thực hiện phép tính :
\((\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}).\dfrac{1-3-5-7-...-49}{89}\)
3.a, Tìm x biết : \(|2x+3|=x+2\)
b, Tìm giá trị nhỏ nhất của \(A=|x-2006|+|2007-x|\) khi x thay đổi
Bài 1:
\(\frac{1}{8}.16^n=2^n\)
\(\Rightarrow\frac{16^n}{8}=2^n\)
\(\Rightarrow\frac{\left(2^4\right)^n}{2^3}=2^n\)
\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)
\(\Rightarrow2^{4n-3}=2^n\)
\(\Rightarrow4n-3=n\)
\(\Rightarrow4n-n=3\)
\(\Rightarrow3n=3\)
\(\Rightarrow n=3:3\)
\(\Rightarrow n=1\left(TM\right).\)
Vậy \(n=1.\)
Bài 3:
a) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-x=2-3\\2x+x=-2-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\left(-1\right):1\\x=\left(-5\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{5}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-1;-\frac{5}{3}\right\}.\)
Chúc bạn học tốt!
Bài 3:
b) \(A=\left|x-2006\right|+\left|2007-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|x-2006+2007-x\right|\)
\(\Rightarrow A\ge\left|1\right|\)
\(\Rightarrow A\ge1.\)
Dấu '' = '' xảy ra khi:
\(\left(x-2006\right).\left(2007-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2006\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2006\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2006\\x\le2007\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2006\\x\ge2007\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2006\le x\le2007\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=1\) khi \(2006\le x\le2007.\)
Chúc bạn học tốt!
Câu 2 :
Đặt : \(P=\left(\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{44.49}\right).\frac{1-3-5-7...-49}{89}\)
\(=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+....+\frac{5.}{44.49}\right).\frac{1-3-5-....-49}{89}\)
\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-3-5-...-49}{89}\)
\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}\)
\(=\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)
\(=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)
Vậy : \(P=-\frac{9}{28}\)
A = \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)