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

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

Do đó: A=B

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

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

--> Bằng nhau

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

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

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

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

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

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

\(5-\sqrt{8}>\sqrt{3}\)

căn 3 < 5 - căn 8

Mik sắp khôi phục đc điểm âm rồi, chỉ còn 11 nữa thôi, các bn giúp mik nha~ 

khi nào hết âm mik sẽ ra câu hỏi và giúp lại các bn!!

Cảm ơn trc~~

căn 3 < 5 - căn 8

làm ơn k cho mình nhé!!!

\(5-\sqrt{8}>5-\sqrt{9}=5-3=2=\sqrt{4}>\sqrt{3}\)

Vậy \(\sqrt{3}< 5-\sqrt{8}\)

\(5-\sqrt{5}.\sqrt{3}=5-\sqrt{5.3}=5-\sqrt{15}\)

\(1=5-4=5-\sqrt{16}\)

-Vì \(-\sqrt{15}>-\sqrt{16}\) nên \(5-\sqrt{15}>5-\sqrt{16}\)

\(\Rightarrow5-\sqrt{5}.\sqrt{3}>1\)

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

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

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)

\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)

Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)

b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)

\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)

Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)

$\sqrt{3}<\sqrt{4}=2$

$5-\sqrt{8}=5-2\sqrt{2}$ mà

$2\sqrt{2}=\sqrt{8}<\sqrt{9}=3$nên $5-2\sqrt{2}>5-3=2$

Vậy $\sqrt{3}<5-\sqrt{8}$

$\sqrt{3}<\sqrt{4}=2$

$5-\sqrt{8}=5-2\sqrt{2}$ mà

2√2=√8<√9=322=8<9=3nFe=nFe2On=a(mol)nên 5−2√2>5−3=25−22>5−3=256a+a(112+16n)=14,4(1)

Vậy √3<5−√8nSO2=0,1(mol)

\(\left(\sqrt{3}\right)^2=3\)

\(\left(5-\sqrt{8}\right)^2=33-10\sqrt{8}=3+30-10\sqrt{8}\)

mà \(0< 30-10\sqrt{8}\)

nên \(\sqrt{3}< 5-\sqrt{8}\)