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 \(\frac{1}{10^{4024}}\)
\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Rightarrow\frac{1}{6}.6.6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x.\left(1+6\right)=7.6^9\)
\(\Rightarrow6^x=\frac{7.6^9}{7}=6^9\)
\(\Rightarrow x=9\)
\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow\frac{1}{6}.6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow6^{x+1}.\left(\frac{1}{6}+1\right)=7.6^9\)
\(\Leftrightarrow6^{x+1}.\frac{7}{6}=7.6^9\)
\(\Leftrightarrow6^{x+1}=7.6^9:\frac{7}{6}\)
\(\Leftrightarrow6^{x+1}=7.6^9.\frac{6}{7}\)
\(\Leftrightarrow6^{x+1}=\left(7.\frac{6}{7}\right).6^9\)
\(\Leftrightarrow6^{x+1}=6.6^9\)
\(\Leftrightarrow6^{x+1}=6^{10}\)
\(\Leftrightarrow x+1=10\)
\(\Leftrightarrow x=9\)