[Toán 8] Rút gọn $ (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam

Áp dụng HĐT đáng nhớ :

\(\left(a-b\right)\left(a+b\right)=a^2-b^2\) . Ta có :

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

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

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

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

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

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

Chúc bạn học tốt !!!

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

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

\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)

\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)

\(=a^2\)

\(\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)\left(x^2+x+1\right).\)

\(=\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)^3.\)

\(=\left(x-1\right)^3+4\left(1-x^2\right)+3\left(x-1\right)^3.\)

\(=\left(x-1\right)^3+3\left(x-1\right)^3+4\left(1-x^2\right)\)

\(=4\left(x-1\right)^3+4\left(1-x^2\right)\)

\(=4\left[\left(x-1\right)^3+\left(1-x^2\right)\right]\)

Lời giải:

Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:

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

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

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

\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

Lời giải:

Áp dụng HĐT đáng nhớ \((a-b)(a+b)=a^2-b^2\). Ta có:

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

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

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

\(=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{16}-1)(3^{16}+1)(3^{32}+1)\)

\(=(3^{32}-1)(3^{32}+1)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)

=(2^4-1)(2^4+1)(2^8+1)(2^16+1)

=(2^8-1)(2^8+1)(2^16+1)

=(2^16-1)(2^16+1)

=2^32-1

2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1

chúc bn hok tốt @_@

\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

a) \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

\(=x^2-4-\left(x^2+x-3x-3\right)\)

\(=x^2-4-x^2-x+3x+3\)

\(=2x-1\)

b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

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

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