Question

[do hoang duong]

tim x nguyen de

can bac 2 cua x+1 chia het cho can bac 2 cua x-3

​GIÚP MÌNH ĐI , MAI PHẢI NỘP BÀI RÙI

AI GIUY MINH LIKE CHO

Answer 1

[Tên tôi là Thành]

Ta co : \(\frac{\sqrt{x+1}}{\sqrt{x-3}}=\frac{\sqrt{x-3+4}}{\sqrt{x-3}}=\frac{\sqrt{x-3}}{\sqrt{x-3}}+\frac{4}{\sqrt{x-3}}\)\(=1+\frac{4}{\sqrt{x-3}}\)

De x nguyen thi \(1+\frac{4}{\sqrt{x-3}}\)nguyen

\(\Rightarrow\)\(\frac{4}{\sqrt{x-3}}\)nguyen\(\Rightarrow\)4 chia het cho \(\sqrt{x-3}\)

\(\Rightarrow\)4\(\in B\)cua \(\sqrt{x-3}\)

\(\Rightarrow\sqrt{x-3}\in\left\{4;-4;2;-2;1;-1\right\}\)

 TH1 : \(\sqrt{x-3}=4\)

\(\Rightarrow x-3=16\Rightarrow x=19\)\(\Rightarrow\)chon

 TH2 : \(\sqrt{x-3}=-4\)  vo ly \(\Rightarrow\) loai

TH3 : \(\sqrt{x-3}=2\Rightarrow x-3=4\Rightarrow x=7\Rightarrow\)chon

TH4 : \(\sqrt{x-3}=-2\Rightarrow\)vo ly \(\Rightarrow\)chon

TH5 : \(\sqrt{x-3}=1\Rightarrow x-3=1\Rightarrow x=4\Rightarrow\)chon

TH6 : \(\sqrt{x-3}=-1\Rightarrow\)vo ly\(\Rightarrow\)loai

Vay x\(\in\){19;7;4}

Answer 2

[Nguyễn Tuấn Minh]

lớp 6 đã học căn bậc đâu