e: \(\left(a^2-1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)

\(=\left(a^3-1\right)\left(a^3+1\right)\)

\(=a^6-1\)

b: Ta có: \(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)\)

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

\(=1-x^6\)

c: \(\left(a+1\right)\left(a+2\right)\left(a^2+4\right)\left(a-1\right)\left(a^2+1\right)\left(a-2\right)\)

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

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

\(=\left(a^4-1\right)\left(a^4-16\right)\)

\(=a^8-17a^4+16\)

d: \(\left(a^3+3\right)\left(a^6-3a^3+9\right)\)

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

\(=a^9+27\)

\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

a, \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{200}-1\right)\)

\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{200}\right)\)

\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{199}{200}\)

\(-A=\frac{1}{200}\)

\(A=\frac{-1}{200}>\frac{-1}{199}\)

a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(A=1-\frac{1}{2^{50}}

Bạn Detective_conan giải đúng đấy!

YÊU TỚ ĐI RỒI TỐ NOI

yêu khiểu gì được

ĐƯỢC CHỨ KHÔNG SAO

a) Đặt M=1/2+1/2²+1/2³+...+1/2¹⁹⁹⁸

=>2M=1+1/2+1/2²+1/2³+...+1/2¹⁹⁹⁷

2M-M=(1+1/2+1/2²+1/2³+...+1/2¹⁹⁹⁷)-(1/2+1/2²+1/2³+...+1/2¹⁹⁹⁸)

M=1-1/2¹⁹⁹⁸

`A=sqrt{1+1/a^2+1/(a+1)^2}`
`=sqrt{1/a^2+2/a+1-2/a+1/(a+1)^2}`
`=sqrt{(1/a+1)^2-2/a+1/(a+1)^2}`
`=sqrt{(a+1)^2/a^2-2.(a+1)/a.(1/(a+1))+1/(a+1)^2}`
`=sqrt{((a+1)/a-1/(a+1))^2}`
`=|(a+1)/a-1/(a+1)|`
`=|1+1/a-1/(a+1)|`
`a>0=>1/a>1/(a+1)=>1+1/a-1/(a+1)>0`
`=>A=1+1/a-1/(a+1)`

Áp dụng công thức ở A ta tính được

`B=1+1/1-1/2+1+1/2-1/3+1-1/3+1/4+.......+1+1/(n-1)-1/n`(ở sau bạn không ghi rõ nên mình đặt số cuối là n)

`=underbrace{1+1+....+1}_{\text{n chữ số 1}}-1/n`

`=n-1/n`

4333344

?reeeeeeeeeeee

Ủa, cái số gì đây??????