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a,\(\sqrt{12}=2\sqrt{3}=\sqrt{3}+\sqrt{3}\)
ta có \(\sqrt{5}>\sqrt{3}\)và\(\sqrt{7}>\sqrt{3}\)=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
a)\(\sqrt{8}+3< \sqrt{9}+3=3+3=6< 6+\sqrt{2}\)
b)\(14=\sqrt{196}>\sqrt{195}=\sqrt{13.15}=\sqrt{13}.\sqrt{15}\)
c) Ta có: \(\hept{\begin{cases}\sqrt{27}>\sqrt{25}=5\\\sqrt{6}>\sqrt{4}=2\end{cases}\Rightarrow\sqrt{27}+\sqrt{6}+1>5+2+1=8}\)
Mà \(\sqrt{48}< \sqrt{49}=7< 8\)
\(\Rightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)
Tham khảo nhé~
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
Cái này mk làm câu a), câu b) bạn tự áp dụng nha :3
a)
\(\sqrt{6}-\sqrt{7}=\frac{\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{6}+\sqrt{7}\right)}{\left(\sqrt{6}+\sqrt{7}\right)}=\frac{6-7}{\sqrt{6}+\sqrt{7}}=\frac{-1}{\sqrt{6}+\sqrt{7}}\)
Tương tự ta có \(\sqrt{7}-\sqrt{8}=\frac{-1}{\sqrt{7}+\sqrt{8}}\)
Dễ dàng thấy \(\sqrt{7}+\sqrt{8}>\sqrt{6}+\sqrt{7}\Rightarrow\frac{1}{\sqrt{7}+\sqrt{8}}< \frac{1}{\sqrt{6}+\sqrt{7}}\Leftrightarrow\frac{-1}{\sqrt{7}+\sqrt{8}}>\frac{-1}{\sqrt{6}+\sqrt{7}}\)
Vậy \(\sqrt{7}-\sqrt{8}>\sqrt{6}-\sqrt{7}\)
Lời giải:
a)
\(\sqrt{6}-\sqrt{7}=\frac{6-7}{\sqrt{6}+\sqrt{7}}=\frac{-1}{\sqrt{6}+\sqrt{7}}\)
\(\sqrt{7}-\sqrt{8}=\frac{7-8}{\sqrt{7}+\sqrt{8}}=\frac{-1}{\sqrt{7}+\sqrt{8}}\)
Thấy rằng \(\sqrt{6}+\sqrt{7}< \sqrt{7}+\sqrt{8}\)
\(\Rightarrow \frac{1}{\sqrt{6}+\sqrt{7}}> \frac{1}{\sqrt{7}+\sqrt{8}}\Rightarrow \frac{-1}{\sqrt{6}+\sqrt{7}}< \frac{-1}{\sqrt{7}+\sqrt{8}}\)
Hay $\sqrt{6}-\sqrt{7}< \sqrt{7}-\sqrt{8}$
b)
\(\sqrt{15}-\sqrt{14}=\frac{15-14}{\sqrt{15}+\sqrt{14}}=\frac{1}{\sqrt{15}+\sqrt{14}}\)
\(\sqrt{13}-\sqrt{12}=\frac{13-12}{\sqrt{13}+\sqrt{12}}=\frac{1}{\sqrt{13}+\sqrt{12}}\)
Dễ thấy \(\sqrt{15}+\sqrt{14}> \sqrt{13}+\sqrt{12}\Rightarrow \frac{1}{\sqrt{15}+\sqrt{14}}< \frac{1}{\sqrt{13}+\sqrt{12}}\)
Hay \(\sqrt{15}-\sqrt{14}< \sqrt{13}-\sqrt{12}\)
Thank you ^.^