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13 tháng 11 2018

\(\left(\sqrt{0,36}.5-\sqrt{\left(0,5\right)^2}\div\frac{1}{6}\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(\left(0,6.5-0,5.6\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(\left(3-3\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(0.\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

= 0

Bài 1:...
Đọc tiếp

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

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

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

10 tháng 3 2020

d) \(\left(-45,7\right)+\left[\left(+5,7\right)+\left(+5,75\right)+\left(-0,75\right)\right]\)

\(=\left(-45,7\right)+\left[5,7+5,75-0,75\right]\)

\(=\left(-45,7\right)+5,7+5,75-0,75\)

\(=\left[\left(-45,7+5,7\right)\right]+\left[5,75-0,75\right]\)

\(=-40+5=-35\)

e) \(11,26-5,13:\left(5\frac{5}{18}-1\frac{8}{9}\cdot1,25+1\frac{16}{63}\right)\)

\(=11,26-5,13:\left(\frac{95}{18}-\frac{17}{9}\cdot\frac{5}{4}+\frac{79}{63}\right)\)

\(=11,26-5,13:\left(\frac{95}{18}-\frac{85}{36}+\frac{79}{63}\right)\)

\(=\frac{563}{50}-\frac{513}{100}:\frac{1051}{252}\)

\(=\frac{563}{50}-\frac{513}{100}\cdot\frac{252}{1051}\)

\(=\frac{563}{50}-\frac{129276}{105100}=\frac{21083}{2102}\)

Số lớn quá!

j) \(\sqrt{8^2+6^2}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)

\(=\sqrt{64+36}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)

\(=\sqrt{100}\cdot4+\frac{1}{2}\cdot\frac{2\sqrt{5}}{5}\)

\(=10\cdot4+\frac{\sqrt{5}}{5}=40+\frac{\sqrt{5}}{5}=\frac{200+\sqrt{5}}{5}\)

h) Cái đây mình có làm rồi

5 tháng 12 2015

câu  nào bạn không làm đc

5 tháng 12 2015

\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)

\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)

\(=\frac{3}{4}-\frac{3}{8}+5\)

\(=\frac{3}{8}+5=\frac{43}{8}\)

\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)

\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)

\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)

\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)

1 tháng 2 2020

\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)