câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm

1) So sánh:

N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)

M = \(\sqrt{18}-\sqrt{8}\)

\(=3\sqrt{2}-2\sqrt{2}\)

\(=\sqrt{2}\)

Ta có: \(1=\sqrt{1}\)

Mà 1 < 2

\(\Rightarrow\sqrt{1}< \sqrt{2}\)

Hay 1 \(< \sqrt{2}\)

Vậy N < M
 

2) Với \(x>0;x\ne4;x\ne9\), ta có:

A = \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

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

\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-4-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x-3}\right)}\)

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

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

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

Đặt: 

\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)

\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)  

\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

Mà: \(4\sqrt{5}>1\)

Nên: \(A^2< B^2\)

\(\Rightarrow A< B\)

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

=>A^2=(căn 10)^2=10=9+1

Đặt B=2+căn 5

=>B^2=(2+căn 5)^2=9+4căn 5

1<4căn 5

=>9+1<9+4căn 5

=>A^2<B^2

=>A<B

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\Rightarrow A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=6+2\sqrt{9-5}=6+2.2=10\)

\(B=2+\sqrt{5}\Rightarrow B^2=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

\(>9+1=10=A^2\)

\(\Rightarrow B^2>A^2\Rightarrow B>A\)

Vậy, B>A

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)

\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

Do đó: A=B

\(\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}=\left|\sqrt{5}+1\right|-\sqrt{5}=1\)

\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

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