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\(4\cdot5^2-3\cdot2^3+3^3\cdot3^2\)
\(=4\cdot25-3\cdot8+3^{2+3}\)
\(=100-24+3^5\)
\(=100-24+243\)
\(=343-24\)
\(=319\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
`#3107.101107`
a,
`2x + 3x = 5^7 \div 5^5`
$\Rightarrow (2 + 3)x = 5^{7 - 5}$
$\Rightarrow 5x = 5^2$
$\Rightarrow 5x = 25$
$\Rightarrow x = 25 \div 5$
$\Rightarrow x = 5$
Vậy, `x = 5`
b,
`5(7x - 45) = 2^3 . 5^2 - 3^2 . 20`
`\Rightarrow 5(7x - 45) = 8.25 - 9.20`
`\Rightarrow 5(7x - 45) = 40.5 - 36.5`
`\Rightarrow 5(7x - 45) = 5.(40 - 36)`
`\Rightarrow 5(7x - 45) = 5.4`
`\Rightarrow 7x - 45 = 4`
`\Rightarrow 7x = 49`
`\Rightarrow x = 49 \div 7`
`\Rightarrow x = 7`
Vậy, `x = 7.`
A) (2+3).x=5 mũ 2
5.x=5 mũ 2
5.x=25
x=25:5
x=5
vậy x=5
câu B ko bt làm bn ạ
a) 5.22 + (x + 3) = 52
5.4 + (x + 3) = 25
20 + (x + 3) = 25
x + 3 = 25 – 20
x + 3 = 5
x = 5 – 3 = 2
b) 23 + (x – 32) = 53 - 43
8 + (x – 9) = 125 – 64
8 + (x – 9) = 61
x – 9 = 61 – 8
x – 9 = 53
x = 53 + 9 = 62
a) \(5.2^2+\left(x+3\right)=5^2\)
\(x+3=5^2-5.2^2\)
\(x+3=25-20\)
\(x+3=5\)
\(x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(8+\left(x-9\right)=125-64\)
\(x-9=53\)
\(x=62\)
\(A=2+2^2+...+2^{20}\)
\(2A=2^2+2^3+...+2^{21}\)
\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)
\(A=2^{21}-2\)
___________
\(B=5+5^2+...+5^{50}\)
\(5B=5^2+5^3+...+5^{51}\)
\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)
\(4B=5^{51}-5\)
\(B=\dfrac{5^{51}-5}{4}\)
___________
\(C=1+3+3^2+...+3^{100}\)
\(3C=3+3^2+...+3^{101}\)
\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)
\(2C=3^{101}-1\)
\(C=\dfrac{3^{101}-1}{2}\)
a) \(S=1+2+2^2+..+2^{2022}\)
\(2S=2+2^2+2^3+...+2^{2023}\)
\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)
\(S=2^{2023}-1\)
b) \(S=3+3^2+3^3+...+3^{2022}\)
\(3S=3^2+3^3+...+3^{2023}\)
\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)
\(2S=3^{2023}-3\)
\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)
c) \(S=4+4^2+4^3+...+4^{2022}\)
\(4S=4^2+4^3+...+4^{2023}\)
\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)
\(3S=4^{2023}-4\)
\(S=\dfrac{4^{2023}-4}{3}\)
d) \(S=5+5^2+...+5^{2022}\)
\(5S=5^2+5^3+...+5^{2023}\)
\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)
\(4S=5^{2023}-5\)
\(S=\dfrac{5^{2023}-5}{4}\)
\(5x+2x\cdot\left(2^3\cdot5-3^2\cdot4\right)+5^2=4^3\)
\(\Rightarrow5x+2x\cdot\left(8\cdot5-9\cdot4\right)+25=64\)
\(\Rightarrow5x+2x\cdot\left(40-36\right)=64-25\)
\(\Rightarrow5x+2x\cdot4=39\)
\(\Rightarrow5x+8x=39\)
\(\Rightarrow x\cdot\left(5+8\right)=39\)
\(\Rightarrow13x=39\)
\(\Rightarrow x=\dfrac{39}{13}\)
\(\Rightarrow x=3\)
Vậy: ...
\(23+5^2-\left\{11-\left[3^2-2.\left(5^2-3.2^3\right)\right]\right\}\\ =23+25-\left\{11-\left[9-2.\left(25-24\right)\right]\right\}\\ =48-\left[11-\left(9-2.1\right)\right]\\ =48-\left(11-7\right)\\ =48-4\\ =44\)