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13+23+33+...+1003
=1+2+1.2.3+3+2.3.4+100+99.100.101
=(1+2+3+...+100)+(1.2.3+2.3.4+...+99.100.101)
=5050+101989800
=101994850
NHỚ T.I.C.K và KB với mk nha
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.1-\left(x-1\right)^2.\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1-1\right)\left(x-1+1\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}}\)
Vậy: \(x\in\left\{1;2;0\right\}\)
\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}.\left[\left(-3\right)^2\right]^{25}-\dfrac{2}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{1}{3}\right)^{50}.\left(-3\right)^{50}-\dfrac{2}{12}\)
\(=\left[\dfrac{1}{3}.\left(-3\right)\right]^{50}-\dfrac{1}{6}\)
\(=\left(-1\right)^{50}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{6}{6}-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
\(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4=\)
=\(\dfrac{1}{3^{50}}\cdot\left(-9\right)^{25}-\dfrac{2}{3}\cdot\dfrac{1}{4}\)
=\(\dfrac{\left(-9\right)^{25}}{9^{25}}-\dfrac{1}{6}\)
=\(-1-\dfrac{1}{6}\)
=\(-\dfrac{6}{6}-\dfrac{1}{6}\)
=\(-\dfrac{7}{6}\)
\(Q=2^3+4^3+...+20^3\)
\(Q=1^3.2^3+2^3.2^3+3^3.2^3+...+10^3.2^3\)
\(Q=\left(1^3+2^3+3^3+...+10^3\right).2^3\)
\(Q=3025.8\)
\(Q=24224\)
C/MINH:
a.106 - 57 chia hết cho 59
Giải
ta có \(10^6-5^7=\left(2\cdot5\right)^6-5^7\)\(=2^6\cdot5^6-5^7=5^6\cdot\left(2^6-5\right)=5^6\cdot59⋮59\)
2n+3+3n+1+2n+3+2n+2
=2n.23+3n.3+2n.23+2n.22
=2n(23+23)+3n.3+2n.22
=2n.24+3n.3+2n.22
=2n(24+22)+3n.3
=2n.20+3n.3
Ủng hộ mk lên 200 điểm với!
Cảm ơn nhìu!
moi hok lop 6