Ta có:

( 5 2 - 1).P = ( 5 2  – 1).12.( 5 2  + 1)( 5 4  + 1)( 5 8  + 1)( 5 16  + 1)

= 12.(  5 2  – 1).( 5 2  + 1)( 5 4 + 1)( 5 8  + 1)( 5 16 + 1)

= 12.(  5 4  - 1)(  5 4  + 1)(  5 8  + 1)( 5 16  + 1)

= 12.(  5 8  - 1)(  5 8  + 1)( 5 16  + 1)

= 12.(  5 16  - 1)( 5 16  + 1)

= 12.(  5 32  - 1)

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????????

\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\left(5^{128}-1\right)=2.5^{128}-2\)

c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{128}-1\right)\)

\(=2\cdot5^{128}-2\)

BÀi 1:

a: (-243+345)-(257+(-55)-129)

=-243+345-257+55+129

=(-243-257)+(345+55)+129

=-500+400+129

=-100+129

=29

b: \(\left(-2\right)^3-45:\left(-3\right)^2-\left(-2019\right)^0\cdot\left(-1\right)^{2019}\)

\(=-8-\frac{45}{9}-1\cdot\left(-1\right)\)

=-8-5+1

=-13+1

=-12

c: \(\left\lbrack27\cdot\left(-67\right)+33\cdot\left(-27\right)\right\rbrack:\left(-30\right)\)

\(=\frac{-27\left(67+33\right)}{-30}\)

\(=\frac{9}{10}\cdot100=90\)

d: \(\left(1^2+2^2+\cdots+100^2\right)\left(3^4-9^2\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)\left(81-81\right)\)

=0

e: \(53\cdot47+53^2-2021^0\)

\(=53\left(53+47\right)-1\)

=5300-1

=5299

i: 38+(-21)+(-58)+(-6)

=(38-58)+(-21-6)

=-20-27

=-47

h: (2020+27)-[2020+(-73)]-129

=2020+27-2020+73-129

=100-129

=-29

\(\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}-1\right)\\ ...\\ 2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\\ 2A=3^{128}-1\)

Vậy \(A=\dfrac{3^{128}-1}{2}.\)

Bài 1:

a: \(A=\frac{x^2}{\left(y+1\right)^2}:\frac{2x}{y+1}:\frac{2x}{y+1}\)

\(=\frac{x^2}{\left(y+1\right)^2}\cdot\frac{y+1}{2x}\cdot\frac{y+1}{2x}\)

\(=\frac{x^2}{4x^2}=\frac14\)

b: \(B=\frac{x^2}{\left(y+1\right)^2}:\left(\frac{2x}{y+1}:\frac{2x}{y+1}\right)\)

\(=\frac{x^2}{\left(y+1\right)^2}:\left(\frac{2x}{y+1}\cdot\frac{y+1}{2x}\right)\)

\(=\frac{x^2}{\left(y+1\right)^2}:1=\frac{x^2}{\left(y+1\right)^2}\)

Bài4:

=>x(x^2+1)=0

>x=0

Bài 5: 

=>\(3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

Lời giải:
a.

$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$

b.

$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$