Tính giá trị của biểu thức
A=\(\left(99-\frac{3^1}{2}\right).\left(189-\frac{3^2}{3}\right).\left(279-\frac{3^3}{4}\right).....\left(181089-\frac{3^{2012}}{2013}\right).\left(181179-\frac{3^{2013}}{2014}\right)\)
\(2\left|2x-6\right|=\dfrac{5}{6}-\left|x-3\right|\)
2:\(\left|x+2013\right|+\left|x+2014\right|+\left|x+2045\right|=2\)
3:\(\left|2x-1\right|=\left|x+1\right|\)
4:\(\sqrt{\left(x+\sqrt{5}\right)}+\sqrt{\left(y-\sqrt{3}\right)^2}+\left|x-y-z\right|=0\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)
Cho \(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2013}+\left(\frac{3}{4}\right)^{2014}\)
Chứng minh A không phải là số nguyên
tìm x y z
a , | x - 1 | + |3 - x | = 2x - 1
b , \(\left|x^2+x+1\right|=x^2+2\)2
c , \(\left(x+1\right)^{30}+\left|y+2\right|+\left|x^2+z\right|=0\)
d , \(\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}\right).x=\frac{2013}{1}+\frac{2012}{2}+.....+\frac{1}{2013}\)
e , \(\left|\left(x+2\right).\left(x^2-15\right)\right|=x+2\)
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2013}+\left(\frac{1}{2}\right)^{2014}\)
Chứng minh A< 1
thuc hien phep tinh
\(\left[6-3.\left(-\frac{1}{3}^2\right)+\sqrt{0,25}\right]:\sqrt{0,\left(9\right)}\)
\(\left(\frac{1}{2}\right)^2.\sqrt{16}-3^2.\sqrt{0,01}+\left(2^2\right)^2-\left(0,0\left(6\right)\right)+\frac{13}{30}\)
C = \(25.\left(-\frac{1}{3}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
D = \(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
E = \(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
F =\(\left(-\frac{3}{2}\right)+|-\frac{5}{6}|-1\frac{1}{2}:6\)
G = \(\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)