a. x(x-5)-4x+20=0

\(\Leftrightarrow\)x(x-5)-4(x-5)=0

\(\Leftrightarrow\)(x-4)(x-5)=0

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}}\)

b, x(x+6)-7x-42=0

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x+6\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}}\)

3, x^3-5x^2+x-5=0

\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)

a, \(3x^2-6x=3x\left(x-2\right)\)

b, \(4x^2-8x+4=\left(2x\right)^2-2.2.2x+2^2=\left(2x-2\right)^2\)

c, \(2x^2-2=2\left(x^2-1\right)=2\left(x-1\right)\left(x+1\right)\)

d, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)

\(\left[5-\left(2x+y\right)\right]\left(5+2x+y\right)=\left(5-2x+y\right)\left(5+2x+y\right)\)

e, \(8x^3-\frac{1}{27}=\left(2x\right)^3-\left(\frac{1}{3}\right)^3=\left(2x-\frac{1}{3}\right)\left[\left(2x\right)^2+2x.\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]=\left(2x-\frac{1}{3}\right)\left(4x^2+\frac{2x}{3}+\frac{1}{9}\right)\)

a) \(3x^2-6x=3x\left(x-2\right)\)

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

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

d) \(25-4x^2-4xy-y^2=-\left(4x^2+4xy+y^2-25\right)\)

\(=-\left[\left(2x+y\right)^2-5^2\right]=-\left(2x+y+5\right)\left(2x+y-5\right)\)

e) \(8x^3-\frac{1}{27}=\left(2x\right)^3-\left(\frac{1}{3}\right)^3=\left(2x-\frac{1}{3}\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)\)

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

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

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