a)  (x+3)(x+5)=0

=>x+3=0 hoặc x+5=0

=>x=-3 hoặc -5

b) (x-1).5-1=0

=>5x-5-1=0

=>5x-6=0

=>5x=6

=>x=6/5

c) 

làm câu c,d,b và E đi bạn

OK bạn

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)

⇔ \(\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1+\dfrac{x+4}{2017}+1=0\)

\(\Leftrightarrow\) \(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}=0\)

⇔ \(\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)

\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)

⇒ \(x+2021=0\)

⇔ \(x=-2021\)

\(Vậy\) \(x=-2021\)

    \(x+y+z=0\)

\(\Leftrightarrow\)\(\left(x+y+z\right)^2=0\)

\(\Leftrightarrow\)\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\)\(x^2+y^2+z^2=0\)   (vì  xy + yz + xz = 0)

\(\Rightarrow\)\(x=y=z=0\)

Vậy   \(Q=\left(x-1\right)^{2018}+\left(y-1\right)^{2019}+\left(z-1\right)^{2020}=1\)

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3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">3 là sẽ tìm được nghiệm nguyên của PT