Mình làm câu c trước để bạn hình dung ra nhé, câu a tương tự:

c) \(7\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)

\(=\left(8-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)

\(=\left[\left(2^3-1\right)\left(2^3+1\right)\right]\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)

\(=\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)

\(=\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)

\(=\left(2^{12}-1\right)\left(2^{24}+1\right)\)

\(=2^{36}-1\)

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

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

\(=x^8.\left(-4\right)\)

\(a,\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

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

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

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

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

\(=2^{16}-1\)

Ta có \(2ab+b^2+a^2-c^2\)

= \(\left(a+b\right)^2-c^2\)

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

= \(4m\left(a+b-c\right)\)(1)

và \(a+b+c=4m\)<=> \(a+b=4m-c\)(2)

Thế (2) vào (1), ta có:

\(4m\left(4m-c-c\right)=4m\left(4m-2c\right)=16m^2-8mc\)

Vậy \(2ab+b^2+a^2-c^2=16m^2-8mc\)(đpcm)

a,\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b,Áp dụng câu a:

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

\(=\frac{1}{x}\)

Tương thẳng cô-si 3 số cho giả thiết và cái gt đi,t dùng đt ko làm đc