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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)
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 \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)
\(7^2=49=7+42\)
mà \(15+2\sqrt{105}< 42\)
nên \(\sqrt{7}+\sqrt{15}< 7\)
b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)
\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)
mà \(2\sqrt{22}< 15+10\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: N=6-\(\sqrt{11}\)
=\(\sqrt{36}\)-\(\sqrt{11}\)
ta có \(\sqrt{31}\)<\(\sqrt{36}\);\(\sqrt{13}\)>\(\sqrt{11}\)
\(\Rightarrow\)\(\sqrt{31}\)-\(\sqrt{13}\)<\(\sqrt{36}\)-\(\sqrt{11}\)
\(\Leftrightarrow\)\(\sqrt{31}\)-\(\sqrt{13}\)<6-\(\sqrt{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
giải thích giùm mình với
bn giải thích đi rồi mik tích đúng cho
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x=2\sqrt{7}=\sqrt{28}>\sqrt{27}=y\)
b: \(x=6\sqrt{2}=\sqrt{72}< \sqrt{75}=y\)
Mít cứ bình phương lên là ok
(2\(\sqrt{7}\))2 =28 (1)
(3\(\sqrt{3}\))2 =27 (2)
vậy (1) > (2)
cứ thế mà làm là hết mít