\(1-3+3^2+3^3+...+\left(-3\right)^x=\frac{9^{1006-1}}{4}\)
\(1-3+3^2+3^3+.....+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
Tìm x
Tìm x nguyên biết
\(1-3+3^2-3^3+.....+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
tìm so nguyen x biet: a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+..........+\frac{1}{\left(2x-1\right)x\left(2x+1\right)}=\frac{49}{99}\)
b) 1-3+32-33+.........+(-3)x=\(\frac{9^{1006}-1}{4}\)
Tìm x nguyên biết :
\(1-3+3^2-3^3+...+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
\(1-3+3^2-3^3+.........+\left(-3\right)^x=\frac{9^{1006}-1}{4}\)
Tìm x
ai giải giúp mình bài này mình tích cho
\(1-3+3^2-3^3+......+\left(-3\right)^2=\frac{9^{1006}-1}{4}\)
Tìm x
Cần ghấp ai giải đúng tớ tick cho
tìm x, x\(\in\)Q:
a) \(\frac{\left(x+\frac{3}{4}\right).\frac{7}{2}-\frac{1}{6}}{-\left(\frac{4}{5}+\frac{1}{3}\right).\frac{1}{2}+1}=2\frac{33}{52}\)
b) \(\frac{\left(5-\frac{2}{7}\right).\frac{7}{9}:\frac{3}{5}}{\left(3x-\frac{5}{6}\right):\frac{1}{7}}=5\frac{5}{21}\)
BT2: Tim x
\(1.\left(3x+1\right)^2=25\)
\(2.\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)
\(3.\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)
\(4.\frac{1}{9}.3^4.3x=3^7\)
\(5.\frac{x}{5}=\frac{4}{21}\)