\(B=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

    \(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

    \(=5^2-2.5+10\)

      \(=25\)

  B = (3 + 1).(3² + 1).(3⁴ + 1).(3⁸ + 1).(3¹⁶ + 1)

2B = (3 - 1).(3 + 1).(3² + 1).(3⁴ + 1).(3⁸ + 1).(3¹⁶ + 1)

     = (3² - 1).(3² + 1).(3⁴ + 1).(3⁸ + 1).(3¹⁶ + 1)

     = (3⁴ - 1).(3⁴ + 1).(3⁸ + 1).(3¹⁶ + 1)

     = (3⁸ - 1).(3⁸ + 1).(3¹⁶ + 1)

     = (3¹⁶ - 1).(3¹⁶ + 1)

     = 3³² - 1

Vậy A = B

lộn r` bn B = 3³²-1 / 2

Vậy A > B

\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

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

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

     \(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

       \(=\frac{1}{2}\left(5^{32}-1\right)\)

Ta có:   \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

     \(\Leftrightarrow P=\frac{\left(5^2-1\right)}{2}\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

     \(\Leftrightarrow P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

      \(\Leftrightarrow P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

      \(\Leftrightarrow P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

       \(\Leftrightarrow P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

       \(\Leftrightarrow P=\frac{5^{32}-1}{2}\)

Vậy \(P=\frac{5^{32}-1}{2}\)

2x² + 3 ( x - 1 ) ( x - 1 ) = 5x( x + 1 )

=>2x²+3x²-6x+3=5x²+5x

=>5x²-6x+3=5x²+5x

=>(5x²-5x²)-6x-5x=0-3

=>-11x=-3

=>x=3/11

Xét vế trái:

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3a^2bc+3abc^2+c^3-a^3-b^3-c^3\)

\(=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3\)

\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Vậy: \(\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

(Nhớ k cho mình với nhá!)

\(=39^{13}\left(39^7+1\right)\)

\(=39^{13}.\left(39^7+1^7\right)\)

\(=39^{13}.\left(39+1\right).A\)

\(=40.39^{13}.A\)chia hết cho 40