\(\frac{2}{ab}-9=\frac{1}{c^2}\)\(\Rightarrow\frac{2}{ab}-\frac{1}{c^2}=9\)

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\left(\frac{2}{ab}-\frac{1}{c^2}\right)=3^2-9\)

\(\Rightarrow\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2.\frac{1}{a}.\frac{1}{b}+2.\frac{1}{b}.\frac{1}{c}+2.\frac{1}{c}.\frac{1}{a}-\frac{2}{ab}+\frac{1}{c^2}=0\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}-\frac{2}{ab}+\frac{1}{c^2}=0\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ac}+\frac{1}{c^2}=0\)

\(\Rightarrow\left(\frac{1}{a^2}+\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}\right)=0\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{c}=0\\\frac{1}{b}+\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{-1}{c}\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)\(\Rightarrow\frac{-1}{c}+\frac{-1}{c}+\frac{1}{c}=3\)\(\Rightarrow\frac{-1}{c}=3\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=3\)\(\Rightarrow c=-\frac{1}{3}\)và\(a=b=\frac{1}{3}\)

Lại có: \(P=\left(a+3b+c\right)^{2020}=\left(\frac{1}{3}+3.\frac{1}{3}+\frac{-1}{3}\right)^{2020}=1^{2020}=1\)

x²⁰¹⁷+x²⁰¹⁸+1

=x²⁰¹⁷.(x+x²)+1

=>x²⁰¹⁷.(x+x²)\(⋮\)x²+x

Mà 1\(⋮\)1

=>x²⁰¹⁷.(x+x²)+1\(⋮\)x²+x+1

Đây là cách nghĩ của em ,em ms lớp 6 nên sai sót j a đừng tích sai e nha

Chúc a học tốt

\(x^{2017}+x^{2018}+1=\left(x^{2016}+x^{2017}+x^{2018}\right)-\left(x^{2016}-1\right)\)

\(=x^{2016}\left(x^2+x+1\right)-\left(x^{2016}-1\right)\)

Ta có: \(x^{2016}-1=x^{3.672}-1=\left(x^3\right)^{672}-1^{672}⋮\left(x^3-1\right)⋮\left(x^2+x+1\right)\)

mà \(x^{2016}\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow x^{2017}+x^{2018}+1⋮\left(x^2+x+1\right)\)

ĐKXĐ....

\(\Leftrightarrow4x^2-5x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x^3-3x+x^2-2x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)^2=-2\sqrt{x-1}\)

Đặt \(\sqrt{x-1}=t\left(t\ge0\right)\Rightarrow t^2=x-1\Leftrightarrow x=t^2+1\)

\(\Leftrightarrow3\left(t^2+1\right)t^2+t^4=-2t\)

\(\Leftrightarrow3t^4+3t^2+t^4+2t=0\)

\(\Leftrightarrow4t^4+3t^2+2t=0\)

\(\Leftrightarrow t\left(4t^3+3t+2\right)=0\Leftrightarrow\orbr{\begin{cases}t=0\\t=-\frac{1}{2}\left(l\right)\end{cases}}\)

\(\Rightarrow t=0\Leftrightarrow x=1\left(tm\right)\)

Vậy x =1 

  \(Ta\) \(có :\)\(x^2+91=y^2\)

\(\Rightarrow\)\(x^2 - y^2 = - 91\)

\(\Rightarrow\)\(( x - y)(x +y)=-91\)

\(Ta\)  \(Lập\)  \(Bảng :\)

\(Vậy :..............\)

Mki lp 7 nên ko bít làm! Sorry cậu nha

Theo đề bài: x + 1/x = a => (x + 1/x) = a^2 => x^2 + 1/x^2 = a^2 - 2
=> (x^2 + 1/x^2)^2 = (a^2 - 2)^2
=> x^4 + 1/x^4 + 2 = a^4 - 4a^2 + 4
=> x^4 + 1/x^4 = a^4 - 4a^2 + 2
Sử dụng hằng đẳng thức, ta có:
m^5 + n^5 = (m + n)(m^4 - m^3n + m^2n^2 - mn^3 + n^4)
Áp dụng, ta có:
x^5 + 1/x^5 = (x + 1/x)(x^4 - x^3.(1/x) + x^2.(1/x^2) - x.(1/x^3) + 1/x^4)
= (x + 1/x)(x^4 - x^2 + 1 - 1/x^2 + 1/x^4)
= (x + 1/x)(x^4 + 1/x^4 - (x^2 + 1/x^2) + 1)
= a(a^4 - 4a^2 + 2 - (a^2 - 2) + 1)
= a(a^4 - 4a^2 + 2 - a^2 + 2 + 1)
= a^5 - 5a^3 + 5a

https://h.vn/hoi-dap/question/66209.html " bài này a e gặp r nè " nên e gửi link qua cho chị

ok gửi qua tin nhắn

https://olm.vn/thanhvien/chibiverycute con lồn này bố láo òm

https://olm.vn/thanhvien/chibiverycute con lồn này bố láo òm