4².120-4³.17+4².34

=4².120-4².4.17+4².34

=4².120-4².68+4².34

=4².(120-68+34)

=16.86

=1376

=4².120-4².4¹.17+4².34

=4².120-4².68+4².34

=4².(120-68+34)

=16.86

1376

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\(\dfrac{-4}{99}\)

-4/99

\(\dfrac{2^{39}+2^{35}}{2^{33}\cdot4}\)

\(=\dfrac{2^{35}\cdot\left(2^4+1\right)}{2^{33}\cdot2^2}\)

\(=\dfrac{2^{35}\cdot\left(16+1\right)}{2^{35}}\)

\(=17\)

\(---\)

\(\dfrac{4^{17}+4^3}{4^{16}+4^2}\)

\(=\dfrac{4^3\cdot\left(4^{14}+1\right)}{4^2\cdot\left(4^{14}+1\right)}\)

\(=4\)

#\(Toru\)

\(4\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{79.82}\right)\)

\(=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{79.8}\right)\)

\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{79}-\frac{1}{82}\right)\)

\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{82}\right)\)

\(=\frac{4}{3}.\frac{39}{164}=\frac{13}{41}\)

1: =-3x5+2x5-7x5+7x4=-8x5+7x4=-40+28=-12

2: =-3x5+3x6-4x3+4x2=-9+8=-1

3: =17(-84-16)=-1700

4: =1975(-115+15)=-197500

5: =-145x13+145x57+57x10-57x145=-1315

1/ \(5\left(-3+2\right)-7\left(5-4\right)\)

\(=5.\left(-3\right)+5.2-7.5-7.\left(-4\right)\)

\(=-15+10-35+28\)

\(=-5-7\)

\(=-12\)

2/ \(3.\left(-5+6\right)-4.\left(3-2\right)\)

\(=3.\left(-5\right)+3.6-4.3-4.\left(-2\right)\)

\(=-15+18-12+8\)

\(=3-4\)

\(=-1\)

3/ \(17.\left(-84\right)+17.\left(-16\right)\)

\(=17.\left(-84-16\right)\)

\(=17.\left(-100\right)\)

\(=-1700\)

4/ \(1975.\left(-115\right)+1975.15\)

\(=1975.\left(-115+15\right)\)

\(=1975.\left(-100\right)\)

\(=-197500\)

5/ \(-145.\left(13-57\right)+57.\left(10-145\right)\)

\(=-145.13-145.\left(-57\right)+57.10+57.\left(-145\right)\)

\(=-145.\left(-57+57\right)-145.13+57.10\)

\(=0-1885+570\)

\(=-1315\)

6/ \(157.17-157.7\)

\(=157.\left(17-7\right)\)

\(=157.10\)

\(=1570\)

7/ \(199.\left(15-17\right)-199.\left(-17+5\right)\)

\(=199.\left(-2\right)-199.\left(-12\right)\)

\(=199.\left(-2\right)+199.12\)

\(=199.\left(-2+12\right)\)

\(=199.10\)

\(=1990\)

8/ \(-39.\left(5-99\right)+99.\left(10-39\right)\)

\(=-39.5-39.\left(-99\right)+99.10+99.\left(-39\right)\)

\(=-39.\left(-99+99\right)-39.5+99.10\)

\(=0-195+990\)

\(=795\)

9/ \(-38.\left(25-4\right)+25.\left(-4+38\right)\)

\(=-38.25-38.\left(-4\right)+25.\left(-4\right)+25.38\)

\(=25.\left(-38+38\right)-4.\left(-38+25\right)\)

\(=0-4.\left(-13\right)\)

\(=52\)

10/ \(\left(-37\right).86+37.76\)

\(=37.\left(-86\right)+37.76\)

\(=37.\left(-86+76\right)\)

\(=37.\left(-10\right)\)

\(=-370\)

\(\left(1-\frac{4}{36}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{625}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac27\right)\cdot\ldots\cdot\left(1-\frac{2}{25}\right)\left(1+\frac26\right)\left(1+\frac27\right)\cdot\ldots\cdot\left(1+\frac{2}{25}\right)\)

\(=\frac46\cdot\frac57\cdot\ldots\cdot\frac{23}{25}\cdot\frac86\cdot\frac97\cdot\ldots\cdot\frac{27}{25}\)

\(=\left(\frac46\cdot\frac68\cdot\ldots\cdot\frac{22}{24}\right)\left(\frac57\cdot\frac79\cdot\ldots\cdot\frac{23}{25}\right)\cdot\left(\frac86\cdot\frac{10}{8}\cdot\ldots\cdot\frac{26}{24}\right)\cdot\left(\frac97\cdot\frac{11}{9}\cdot\ldots\cdot\frac{27}{25}\right)\)

\(=\frac{4}{24}\cdot\frac{5}{25}\cdot\frac{26}{6}\cdot\frac{27}{7}=\frac14\cdot\frac15\cdot\frac{13}{3}\cdot\frac{27}{7}=\frac{13\cdot9}{7\cdot4\cdot5}\)

\(\frac{1}{23}-\frac{4}{23\cdot25}\)

\(=\frac{25-4}{25\cdot23}=\frac{21}{25\cdot23}\)

Ta có: \(C=\left(\frac{1}{23}-\frac{4}{23\cdot25}\right)\cdot\left(\left(1-\frac{4}{36}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{625}\right)\right)\)

\(=\frac{7\cdot3}{5\cdot5\cdot23}\cdot\frac{13\cdot9}{7\cdot4\cdot5}=\frac{3\cdot13\cdot9}{5\cdot5\cdot23\cdot4\cdot5}=\frac{27\cdot13}{125\cdot4\cdot23}=\frac{351}{11500}\)