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9 tháng 9 2016

a)n=1

b)n=4

c)n=1

d)n=6

e)n=-1

18 tháng 6 2016

\(a,9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^{\left(-4\right)}.3^2=3^{2+3-4+2}=3^3.\)

\(b,4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.2^{-4}\right)=2^{2+5}:2^{3-4}=2^7:2^{-1}=2^{7-\left(-1\right)}=2^8.\)

\(c,3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=\left(\frac{3^2}{3^2}\right).\left(2^5.2^2\right)=1.2^{5+2}=2^7\)

\(d,\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^2.\frac{1}{3}.\left(3^2\right)^2=\left(\frac{1}{3}\right)^{2+1}.3^4=\left(\frac{1}{3}\right)^3.\left(\frac{1}{3}\right)^{-4}=\left(\frac{1}{3}\right)^{3-4}=\left(\frac{1}{3}\right)^{-1}=3^1\)

Bài 1:...
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Bài 1: Tính

a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)

b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)

c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)

Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)

b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\) 

c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)

d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)

e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)

Bài 3: Chứng minh rằng

a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)

b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)

Bài 4: 

a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)

b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)

c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)

1

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

1 tháng 9 2019

a, Tự chép đề bài ((:

\(=\frac{1}{9}\cdot1+\left(-\frac{1}{243}\right)\cdot\frac{9}{2}\)

\(=\frac{1}{9}-\frac{1}{54}\)

\(=\frac{5}{54}\)

b, 1. \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15}\)

\(=\left(\frac{2^5}{2^4\cdot2^4}\right)^5=\left(\frac{2^5}{2^8}\right)^5=\left(\frac{1}{2^3}\right)^5=\left(\frac{1}{8}\right)^5=\frac{1}{8^5}\)(Để vậy đi :v)

     2. \(\left(\frac{2^6}{16^2}\right)^{10}\)

\(=\left(\frac{2^6}{2^8}\right)^{10}=\left(\frac{1}{2^2}\right)^{10}=\frac{1}{2^{20}}\)

c, \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)

\(=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^2}{1}=3^2=9\)