Question

  làm câu II giúp e với ạ

Answer 1

1) Thay x= 9 vào Q ta được : 

√9 + 1 / √ 9 - 2 = 3 + 1 / 3 - 2 = 4  

 

Answer 2

1) Thay x= 9 vào Q ta được : 

√9 + 1 / √ 9 - 2 = 3 + 1 / 3 - 2 = 4 

 

Answer 3

1) thay x = 9 vào Q ta được

\(\dfrac{\sqrt{9}+1}{\sqrt{9}-2}=\dfrac{3+1}{3-2}=\dfrac{4}{1}=4\)

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

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

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

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

để M đạt giá trị nguyên thì [√(x) - 1] / [√(x) + 1] đạt giá trị nguyên

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

=> √(x) + 1 ∈ Ư(2) = {±1; ±2}

\(\sqrt{x}+1=1\Rightarrow x=0\left(TM\right)\\ \sqrt{x}+1=-1\Rightarrow\left(\text{vô nghiệm}\right)\\ \sqrt{x}+1=2\Rightarrow x=1\left(KTM\right)\\ \sqrt{x}+1=-2\Rightarrow\text{vô nghiệm}\)

vậy để M nhận giá trị nguyên thì x = 0

Answer 4

Câu II:

1: Thay x=9 vào Q, ta được:

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

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

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

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

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

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

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

Để M là số nguyên thì \(2\sqrt{x}⋮\sqrt{x}+1\)

=>\(2\sqrt{x}+2-2⋮\sqrt{x}+1\)

=>\(-2⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;2\right\}\)

=>\(\sqrt{x}\in\left\{0;1\right\}\)

=>\(x\in\left\{0;1\right\}\)​