Question

Tìm x nguyên để \(\sqrt{x}+1⋮\sqrt{x}-3\)

Answer 1

duocwj

Answer 2

\(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)

\(\Rightarrow4⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)\)

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

\(\Rightarrow\sqrt{x}\in\left\{2;4;1;5;-1;7\right\}\) 

=> x thuộc {4;16;1;25;1;49}

Answer 3

\(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)+4⋮\sqrt{x}-3\)

\(\Rightarrow4⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\in\)Ư(4)={-1,-2,-4,1,2,4}

<=> \(x\in\left\{4,1,16,25,49\right\}\)

Answer 4

\(\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}-3\right)\)

=> \(\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)⋮\left(\sqrt{x}-3\right)\)

=>\(\sqrt{x}+1-\sqrt{x}+3⋮\sqrt{x}-3\)

=> \(4⋮\sqrt{x}-3\)

=> \(\sqrt{x}-3\inƯ\left(4\right)\)={ 1;-1;2;-2;4;-4}

Ta có bảng sau :

\(\sqrt{x}-3\)	1	-1	2	-2	4	-4
\(\sqrt{x}\)	4	2	5	1	7	-1
x	16	4	25	1	49	1

Answer 5

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

=> \(4⋮\left(\sqrt{x}-3\right)\)=> \(\sqrt{x}-3\inƯ\left(4\right)\)

=> \(\sqrt{x}-3\in\left\{1;-1;4;-4\right\}\)

=> \(\sqrt{x}\in\left\{4;2;7;-1\right\}\)

=>\(x\in\left\{16;4;49;\varnothing\right\}\)

Answer 6

Sơ sót chỗ này

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

=> \(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

=> \(x\in\left\{16;4;25;1;49;\varnothing\right\}\)