K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

8 tháng 12 2020

Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

\(x^2+x=x\left(x+1\right)\)

\(x^2+x+1=x^2+x+1\)

MTC : \(x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

Quy đồng :  

\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)

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

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

8 tháng 12 2020

\(\frac{x}{x^3-1};\frac{x+1}{x^2+x};\frac{x-1}{x^2+x+1}\)

Ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

         \(x^2+x=x\left(x+1\right)\)

        \(x^2+x+1=x^2+x+1\)

\(\Rightarrow MTC=x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

Quy đồng:

\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)

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

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

8 tháng 12 2020

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

\(=\left(-\frac{x+2}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{2-x}{x+2}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{x^2-4x-4-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-3x^2-4x-4+x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}.\frac{-x^2\left(x-2\right)}{x\left(x-3\right)}=\frac{2x^4+8x^2}{x\left(x+2\right)\left(x-3\right)}\)

8 tháng 12 2020

Ta có a + b + 8 = 0

=> x3 + 3x2 + 6x + y3 + 3y2 + 6y + 8 = 0

=> (x3 + 3x2 + 3x + 1) + (y3 + 3y2 + 3y + 1) + (3x + 3y + 6) = 0

=> (x + 1)3 + (y + 1)3 + 3(x + y + 2) = 0

=> (x + y + 2)[(x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3] = 0

Vì (x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3 \(>0\forall x;y\)

=> x + y + 2 = 0

=> x + y = -2

Vậy A = -2

8 tháng 12 2020

xyz bạn ơi! tại sao từ dòng 3 lại thành dòng 4 vậy

thank you bạn!!! <3