5.7^2-24:23
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HN
1
20 tháng 3 2020
a) Ta có: \(8\cdot39\cdot125\)
\(=\left(8\cdot125\right)\cdot39\)
\(=1000\cdot39=39000\)
b) Ta có: \(49\cdot77-30^2+23\cdot49\)
\(=49\cdot77+49\cdot23-30^2\)
\(=49\cdot\left(77+23\right)-900\)
\(=49\cdot100-9\cdot100\)
\(=100\cdot\left(49-9\right)\)
\(=100\cdot40=4000\)
c) Ta có: \(\frac{2340}{5\cdot89-\left(125+5\cdot7^2\right)+5\cdot11}\)
\(=\frac{2340}{5\left(89+11\right)-125-5\cdot7^2}\)
\(=\frac{2340}{5\cdot100-5\cdot25-5\cdot49}\)
\(=\frac{2340}{5\left(100-25-49\right)}=\frac{2340}{5\cdot26}=\frac{2340}{130}=18\)
LM
0
31 tháng 1
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
17 tháng 3 2023
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
NT
2
CM
2 tháng 11 2019
Đáp án là A
Ta có: a = 3 2 .5.7 và b = 2 4 .3.7 nên ƯCLN(a, b) = 3.7
24 sao chia 23 duoc
5.72-24:23
=5.49-\(\frac{24}{23}\)
=245-\(\frac{24}{23}\)
=\(\frac{5635}{23}-\frac{24}{23}\)
=\(\frac{5635-24}{23}\)
=\(\frac{5611}{23}\)
=\(243\frac{22}{23}\)