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\(C=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)...+\left(5^{17}+5^{18}+5^{19}+5^{20}\right)\\ C=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)...+5^{17}\left(1+5+5^2+5^3\right)\\ C=5\cdot156+5^5\cdot156+...+5^{17}\cdot156\\ C=156\left(5+5^5+...+5^{17}\right)\\ C=12\cdot13\left(5+5^5+...+5^{17}\right)⋮17\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
\(a,9234:\left[3\cdot3\cdot\left(1+8^3\right)\right]\\ =9234:\left[9\cdot\left(1+512\right)\right]\\ =9234:\left(9\cdot513\right)\\ =9234:4617\\ =2\\ b,76-\left\{2\cdot\left[2\cdot5^2-\left(31-2\cdot3\right)\right]\right\}+3\cdot25\\ =76-\left\{2\cdot\left[2\cdot25-\left(31-6\right)\right]\right\}+75\\ =76-\left[2\cdot\left(50-25\right)\right]+75\\ =76-\left(2\cdot25\right)+75\\ =76-50+75\\ =26+75\\ =-49\)
a)35-(12-(-14+(-2)))
=35-(12-(-12))
=35-(12+12)
=35-24
=11
b)49-(-54)-23
=49+54-23
=103-23
=80
c)5871:(928-((-82)+247))×5)
=5871:(928-165×5)
=5871:(928-825)
=5871:103
=57
Đặt \(A=1+5^2+5^4+...+5^{40}\)
\(\Rightarrow25A=5^2+5^4+5^6+...+5^{42}\)
Lấy \(25A-A=\left(5^2+5^4+5^6+...+5^{42}\right)-\left(1+5^2+5^4+...+5^{40}\right)\)
\(\Rightarrow24A=5^{42}-1\)
\(\Rightarrow A=\dfrac{5^{42}-1}{24}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(B=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(B=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=A\)
=>A/B=1
E = 1 + 5 + 52 + ... + 519
5E = 5 + 52 + ... + 520
5E - E = (5 + 52 + ... + 520) - (1 + 5 + ... + 519)
4E = 1 - 520
E = \(\frac{1-5^{20}}{4}\)
\(E=1+5+5^2+...+5^{19}\)
nhân 2 vế của \(E\)với 5 ta được :
\(5E=5.\left(1+5+5^2+5^3+...+5^{19}\right)\)
\(5E=5+5^2+5^3+...+5^{20}\)
\(5E-E=\left(5+5^2+5^3+...+5^{20}\right)-\left(1+5+5^2+5^3+...+5^{19}\right)\)
\(4E=5^{20}-1\)
\(E=\left(5^{20}-1\right):4\)
CHÚ Ý: 5E tức là 5.E; 4E tức là 4,E
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