10 \times× 48 | = | 48 \times× | = | 48 | \times× | 10 | = \times× 2 = |
8 \times× 5 | 8 \times× 5 | 5 |
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\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ A=1-\dfrac{1}{2^{100}}\)
\(E=\dfrac{3^2}{2\cdot4}+\dfrac{3^2}{4\cdot6}+...+\dfrac{3^2}{198\cdot200}\\ =3^2\cdot\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{198\cdot200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\dfrac{99}{200}\\ =\dfrac{891}{400}\)
\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)
a) \(\left(165.99+165\right)-\left(163.101-163\right)\)
\(=165\left(99+1\right)-163\left(101-1\right)\)
\(=165.100-163.100\)
\(100\left(165-163\right)\)
\(=100.2\)
\(=200\)
b) \(24.62+48.19\)
\(=24.62+24.2.19\)
\(=24.62+24.38\)
\(=24\left(62+38\right)\)
\(=24.100=2400\)
\((165.99+165)-(163.101-163)\)
\(=(165.99+165.1)-(163.101-163.1)\)
\(=165(99+1)-163(100-1)\)
\(=165.100-163.100\)
\(=(165-163).100=2.100=200\)
\(24.62+48.19\)
\(=24.62+24.2.19 \)
\(=24.62+24.38\)
\(=24(62+38)\)
\(=24.100=2400\)
8 : 8 = 1 16 : 8 = 2 24 : 8 = 3 32 : 8 = 4 40 : 8 = 5 48 : 8 = 6 56 : 8 = 7 64 : 8 = 8 72 : 8 = 9 80 : 8 = 10
8 : 8 = 1
16 : 8 = 2
24 : 8 = 3
32 : 8 = 4
40 : 8 = 5
48 : 8 = 6
56 : 8 = 7
64 : 8 = 8
72 : 8 = 9
80 : 8 = 10
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
Giải:
a) \(\sqrt{a^4.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\left(3-a\right)\right)^2}\)
\(=\left|a^2\left(3-a\right)\right|\)
b) \(\sqrt{27.48.\left(a-3\right)^2}\)
\(=\sqrt{3.9.16.3.\left(a-3\right)^2}\)
\(=\sqrt{3.3.9.16\left(a-3\right)^2}\)
\(=\sqrt{\left(9.4\left(a-3\right)\right)^2}\)
\(=\left|9.4\left(a-3\right)\right|\)
\(=\left|36\left(a-3\right)\right|\)
c) \(\sqrt{48.75a^2}\)
\(=\sqrt{16.3.25.3a^2}\)
\(=\sqrt{\left(4.3.5a\right)^2}\)
\(=\left|4.3.5a\right|\)
\(=\left|60a\right|\)
d) \(\sqrt{2^4.\left(-9\right)^2}\)
\(=\sqrt{2^4.9^2}\)
\(=\sqrt{\left(2^2.9\right)^2}\)
\(=\left|2^2.9\right|\)
\(=\left|36\right|=36\)
Vậy ...
=3628800
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