Tinhs \(\frac{\left(-\frac{5}{7}\right)^n}{\left(-\frac{5}{7}\right)^{n-1}}\) A\(\ge\)1
Tính :
a) \(\frac{\left(\frac{-5}{7}\right)^n}{\left(\frac{-5}{7}\right)^{n-1}}\)( n\(\ge\)1 )
b) \(\frac{\frac{-1}{2}^{2n}}{\left(\frac{-1}{2}\right)^n}\) ( n \(\in\)N )
a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)
b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)
Tính :
a, \(\frac{\frac{\left(-5\right)^n}{\left(7\right)}}{\frac{\left(-5\right)^{n-1}}{7}}\left(n>=1\right)\) b,\(\frac{\frac{\left(-1\right)^{2n}}{2}}{\frac{\left(-1\right)^n}{2}}\left(n\in N\right)\)
Phân số \(\frac{-5}{7}\)và \(\frac{-1}{2}\)nằm trong ngoặc nhưng mình chỉ đóng ngoặc đc tử nên đừng hiểu sai nha
Ai nhanh mình tick
a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)
tính \(\frac{\left(\frac{-5}{7}^{ }\right)n+1^{ }}{\left(\frac{-5}{7}\right)^{ }n}\)(n>=1)
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
Mọi người ơi giải bài tập này hộ tớ đi
Mai tớ kt 1 tiết rồi
a)
\(\frac{\left(2x+1\right)^2}{4}+\frac{\left(2x-1\right)^2}{2}\ge\frac{12\left(x+5\right)^2}{4}\) ;
b)
\(\frac{\left(1-x\right)^2}{7}-\frac{2\left(x+3\right)^2}{3}\le\frac{-11\left(x+5\right)^2}{21}\) ;
c)
\(|5-3x|=2+x\)
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Tính:
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(=\frac{-5}{7}\)
#)Giải :
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\frac{\left(\frac{-5}{7}\right)^n\times\left(\frac{-5}{7}\right)}{\left(\frac{-5}{7}\right)^n}=\frac{-5}{7}\)
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\left(\frac{-5}{7}\right)^{n+1-n}=\left(\frac{-5}{7}\right)^1=\frac{-5}{7}\)