\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}.\)
\(\frac{\left(-1\right)^4}{2^4}=\frac{1}{16}.\)
\(\frac{1}{16}=\frac{1}{16}.\)
Chúc bạn học tốt!
\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}.\)
\(\frac{\left(-1\right)^4}{2^4}=\frac{1}{16}.\)
\(\frac{1}{16}=\frac{1}{16}.\)
Chúc bạn học tốt!
tính: \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
Tính các luỹ thừa sau:
\(\left(0,375\right)^2=\left(\frac{3}{8}\right)^2=\left(\frac{3}{18}\right)^2=\frac{9}{64}\)
Giải giúp mình đi ạ!
Tính P biết :
\(p=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.....+\frac{1}{16}\left(1+2+...+16\right)\)
Tính giá trị biểu thức sau:
\(B=\left(18\frac{1}{3}:\sqrt{225}+8\frac{2}{3}.\sqrt{\frac{49}{4}}\right):\left[\left(12\frac{1}{3}+8\frac{6}{7}\right)-\frac{\left(\sqrt{7}\right)^2}{\left(3.\sqrt{2}\right)^2}\right]:\frac{1704}{445}\)
\(C=\frac{\left(81,624:4\frac{4}{3}-4,505\right)^2+125\frac{3}{4}}{\left\{\left[\left(\frac{11}{25}\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}:\frac{13}{25}}\)
P/s: Giúp mik với ạ! Nguyễn Văn Đạt ; @Vũ Minh Tuấn; @Phạm Thị Diệu Huyền; ..........
Thực hiện phép tính :
a, \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
b, \(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
Tính tổng sau
1) B= 1.2+2.3+3.4+......+99.100
2) C= \(1^2+2^2+3^2+...+99^2\)
3) D= \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
4) E=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
Giúp mik nhé!!
\(\left(9\frac{2}{4}:5,2+3,4\cdot2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)\)
Tìm x, y biết:
\(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{3}\right|+\left|x-\frac{1}{4}\right|+\left|y-\frac{1}{5}\right|=\frac{1}{4}\)
Giúp với ạ T.T
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)