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\(P=\dfrac{Q}{R}=\dfrac{3\sqrt{x}-1}{x-4}:\dfrac{2}{\sqrt{x}-2}\)

\(=\dfrac{3\sqrt{x}-1}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{3\sqrt{x}-1}{2\sqrt{x}+4}\)

P là số nguyên

=>\(3\sqrt{x}-1⋮2\sqrt{x}+4\)

=>\(6\sqrt{x}-2⋮2\sqrt{x}+4\)

=>\(6\sqrt{x}+12-14⋮2\sqrt{x}+4\)

=>\(2\sqrt{x}+4\inƯ\left(-14\right)\)

mà 2*căn x+4>=4

nên \(2\sqrt{x}+4\in\left\{7;14\right\}\)

=>\(2\sqrt{x}\in\left\{3;10\right\}\)

=>\(x\in\left\{\dfrac{9}{4};25\right\}\)

30 tháng 8 2023

Ta có:

\(\dfrac{Q}{R}\) là:

\(\dfrac{3\sqrt{x}-1}{x-4}:\dfrac{2}{\sqrt{x}-2}\) (ĐK: \(x\ge0;x\ne4\))

\(=\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}\right)^2-2^2}:\dfrac{2}{\sqrt{x}-2}\)

\(=\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-1}{2\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-1}{2\sqrt{x}+4}\)

Để giá trị của biểu thức \(\dfrac{Q}{R}\) nguyên thì 

\(3\sqrt{x}-1\) chia hết cho \(2\sqrt{x}+4\)

\(2\left(3\sqrt{x}-1\right)\) chia hết cho \(2\sqrt{x}+4\)

⇒ \(6\sqrt{x}-2\) chia hết cho \(2\sqrt{x}+4\)

⇒ \(6\sqrt{x}+12-14\)  chia hết cho \(2\sqrt{x}+4\)

⇒ \(3\left(2\sqrt{x}+4\right)-14\) chia hết cho \(2\sqrt{x}+4\) 

⇒ - 14 chia hết cho \(2\sqrt{x}+4\)

Mà: Ư(-14)\(=\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

ĐK: \(2\sqrt{x}+4\ge4\)

\(\Rightarrow2\sqrt{x}+4\in\left\{7;14\right\}\)

\(\Rightarrow x\in\left\{\dfrac{9}{4};25\right\}\)

25 tháng 10 2021

\(a,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ b,A=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{1;3\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{0;4\right\}\left(tm\right)\)

25 tháng 10 2021

a) \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(\Rightarrow A=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

1:

a: \(P=B:A\)

\(=\dfrac{3\sqrt{x}-3-2\sqrt{x}+4+\sqrt{x}+2}{x-4}:\dfrac{4\sqrt{x}-1}{x-4}\)

\(=\dfrac{2\sqrt{x}+3}{x-4}\cdot\dfrac{x-4}{4\sqrt{x}-1}=\dfrac{2\sqrt{x}+3}{4\sqrt{x}-1}\)

b: Để P nguyên thì \(2\sqrt{x}+3⋮4\sqrt{x}-1\)

=>\(4\sqrt{x}+6⋮4\sqrt{x}-1\)

=>\(4\sqrt{x}-1+7⋮4\sqrt{x}-1\)

=>4căn x-1 thuộc {1;-1;7;-7}

=>căn x thuộc {1/2;0;2}

mà x nguyên và x>=0 và x<>4

nên x thuộc {0}

11 tháng 7 2023

\(a,A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-5\\ =2\sqrt{2}-2\)

\(b,B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =-\sqrt{7}+3\)

11 tháng 7 2023

a)

\(A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)

\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\) (vì \(2-\sqrt{5}< 0;2\sqrt{2}-\sqrt{5}>0\) )

\(=2\sqrt{2}-2\)

b)

\(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)

\(=2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\) (vì \(\sqrt{7}-2\sqrt{2}< 0;3-2\sqrt{2}>0\) )

\(=3-\sqrt{7}\)

 

26 tháng 1 2022

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right).\dfrac{1}{\sqrt{x}+1}\)

\(\Rightarrow B=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{1}{\sqrt{x}+1}\)

\(\Rightarrow B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\sqrt{x}+1}\)

\(\Rightarrow B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow B=\dfrac{1}{\sqrt{x}}\)

26 tháng 1 2022

Bạn cần làm j với B

8 tháng 12 2021

\(a,B=4\sqrt{x+1}-3\sqrt{x+1}+\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\\ b,B=8\Leftrightarrow4\sqrt{x+1}=8\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\left(tm\right)\)

a: Vì (d)//y=3x nên m-1=3

hay m=2

Vậy: (d): y=3x+2n-5

Thay x=1 và y=4 vào (d), ta được:

2n-5+3=4

\(\Leftrightarrow2n=6\)

hay n=3