a)

\(\begin{array}{l}{\left( {\frac{{ - 1}}{2}} \right)^5} = \frac{{{{\left( { - 1} \right)}^5}}}{{{2^5}}} = \frac{{ - 1}}{{32}};\\{\left( {\frac{{ - 2}}{3}} \right)^4} = \frac{{{{\left( { - 2} \right)}^4}}}{{{3^4}}} = \frac{{16}}{{81}};\\{\left( { - 2\frac{1}{4}} \right)^3} = {\left( {\frac{{ - 9}}{4}} \right)^3} = \frac{{{{\left( { - 9} \right)}^3}}}{{{4^3}}} = \frac{{-729}}{{64}};\\{\left( { - 0,3} \right)^5} = {\left( {\frac{{ - 3}}{{10}}} \right)^5} = \frac{{ - 243}}{{100000}};\\{\left( { - 25,7} \right)^0} = 1\end{array}\)

b)

\(\begin{array}{l}{\left( { - \frac{1}{3}} \right)^2} = \frac{1}{9};\\{\left( { - \frac{1}{3}} \right)^3} = \frac{{ - 1}}{{27}};\\{\left( { - \frac{1}{3}} \right)^4} = \frac{1}{{81}};\\{\left( { - \frac{1}{3}} \right)^5} = \frac{{ - 1}}{{243}}.\end{array}\)

Nhận xét:

+ Luỹ thừa của một số hữu tỉ âm với số mũ chẵn là một số hữu tỉ dương.

+  Luỹ thừa của một số hữu tỉ âm với số mũ lẻ là một số hữu tỉ âm.

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^3} = \frac{{{{\left( { - 2} \right)}^3}}}{{{3^3}}} = \frac{{ - 8}}{{27}};\\{\left( {\frac{{ - 3}}{5}} \right)^2} = \frac{{{{\left( { - 3} \right)}^2}}}{{{5^2}}} = \frac{9}{{25}};\\{\left( { - 0,5} \right)^3} = {\left( {\frac{{ - 1}}{2}} \right)^3} = \frac{{{{\left( { - 1} \right)}^3}}}{{{2^3}}} = \frac{{ - 1}}{8};\\{\left( { - 0,5} \right)^2}=\frac{{{{\left( { - 1} \right)}^2}}}{{{2^2}}} = \frac{{1}}{4};\\\,{\left( {37,57} \right)^0} = 1;\,\\{\left( {3,57} \right)^1} = 3,57.\end{array}\)

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)

\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)

\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)

\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)

\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)

\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)

Lời giải:

a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$

$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$

$=3-\sqrt{7}$

c.

$=|x-3|=x-3$
d.

$=|1-x|=x-1$

$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.

$=\sqrt{(10a)^2}=|10a|=-10a$

3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

=(5^8-1)(5^8+1)(5^16+1)

=(5^16-1)(5^16+1)

=5^32-1

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(C=5^{32}-1\)

4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)

\(D=4^{128}-1\)

5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)

....

\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)

\(E=5^{512}-1\)

\(R=\dfrac{\sqrt{\left(-\dfrac{2}{5}\cdot\dfrac{-5}{8}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-3^3}{4^3}\cdot\dfrac{5^2}{2^6\cdot3^2}\cdot\dfrac{5^4}{3^4}}}\)

\(=\dfrac{\sqrt{\left(\dfrac{1}{4}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-1}{3^3}\cdot\dfrac{25^3}{16^3}}}=\dfrac{5}{8}:\dfrac{-5}{3\cdot4}=\dfrac{5}{8}\cdot\dfrac{3\cdot4}{-5}=-\dfrac{3}{2}\)

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)

\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)

a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)

b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)

A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)

A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)

Rồi tiếp tục làm nhé bạn.

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`