a=b+1

=>a-b=1

Suy ra: VT=(a+b)(a²+b²)(a⁴+b⁴)(a⁸+b⁸)

=(a-b)(a+b)(a²+b²)(a⁴+b⁴)(a⁸+b⁸)

=(a²-b²)(a²+b²)(a⁴+b⁴)(a⁸+b⁸)

=(a⁴-b⁴)(a⁴+b⁴)(a⁸+b⁸)

=(a⁸-b⁸)(a⁸+b⁸)

=a¹⁶-b¹⁶=VP

=>điều phải chứng minh

Sao tự nhiên lại lòi ra số c vậy?

Bất đẳng thức cần chứng minh tương đương:

\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)

\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))

\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).

Vậy ta có đpcm.

\(a^8+b^8-a^6b^2-a^2b^6=\left(a^8-a^6b^2\right)+\left(b^8-a^2b^6\right)=a^6\left(a^2-b^2\right)+b^6\left(b^2-a^2\right)=\left(a^6-b^6\right)\left(a^2-b^2\right)\) nên suy ra được như vậy Quỳnh Anh

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

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

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

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

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

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

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

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

\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

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

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

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

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

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

=(3^4-1)(3^4+1)(3^8+1)(3^16+1)

=(3^8-1)(3^8+1)(3^16+1)

=(3^16-1)(3^16+1)

=3^32-1

2: B=(1-3^2)(1+3^2)*...*(1+3^16)

=(1-3^4)(1+3^4)(1+3^8)(1+3^16)

=1-3^32

1

\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)

\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)

bn cứ quy đồng lần lượt 2 hạng tử đầu tiên là đc thôi

mk giải phần a k ra