Ta so sánh: \(\sqrt{3}-\sqrt{2}\) và \(\sqrt{7}-\sqrt{6}\)

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

\(\sqrt{7}-\sqrt{6}=\frac{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}{\sqrt{7}+\sqrt{6}}=\frac{7-6}{\sqrt{7}+\sqrt{6}}=\frac{1}{\sqrt{7}+\sqrt{6}}\)

Vì \(\sqrt{3}+\sqrt{2}< \sqrt{7}+\sqrt{6}\)

nên \(\frac{1}{\sqrt{3}+\sqrt{2}}>\frac{1}{\sqrt{7}+\sqrt{6}}\)

\(\Rightarrow\sqrt{3}-\sqrt{2}>\sqrt{7}-\sqrt{6}\)

\(\Rightarrow\sqrt{3}+\sqrt{6}>\sqrt{7}+\sqrt{2}\) hay x > y

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

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

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

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

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

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

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

1)\(x>y\)

2)\(x< y\)

3)\(x< y\)

\(a,x=\sqrt{27}-\sqrt{2}\)\(=3\sqrt{3}-\sqrt{2}>3\sqrt{3}-\sqrt{3}=2\sqrt{3}\)

Mà: \(y=\sqrt{3}< 2\sqrt{3}\)

\(\Rightarrow x>y\)

\(b,x=\sqrt{5\sqrt{6}}\Rightarrow x^4=5^2.6=150\)

\(y=\sqrt{6\sqrt{5}}\Rightarrow y^4=6^2.5=180\)

\(\Rightarrow x^4< y^4\Rightarrow x< y\left(x,y>0\right)\)

\(c,x=2m;y=m+2\)

Ta có: \(x-y=2m-\left(m+2\right)=m-2\)

Ta xét các trường hợp:

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

b) \(-6\sqrt[3]{7}=\sqrt[3]{\left(-6\right)^3\cdot7}=\sqrt[3]{-1512}\)

\(7\sqrt[3]{-6}=\sqrt[3]{7^3\cdot\left(-6\right)}=\sqrt[3]{-2058}\)

mà -1512>-2058

nên \(-6\sqrt[3]{7}>7\cdot\sqrt[3]{-6}\)

\(x^2=3+5+2\sqrt{15}=8+\sqrt{60}\)

\(y^2=2+6+2\sqrt{12}=8+\sqrt{48}\)

Mà \(60>48\Rightarrow\sqrt{60}>\sqrt{48}\Rightarrow8+\sqrt{10}>8+\sqrt{48}\)

\(\Rightarrow x^2>y^2\Rightarrow x>y\) (do x;y đều dương)