So sánh:
\(\sqrt{8}+3\)và \(6+\sqrt{2}\)
\(14\)và \(\sqrt{13}.\sqrt{15}\)
\(\sqrt{27}+\sqrt{6}+1\) và \(\sqrt{48}\)
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a,\(\sqrt{12}=2\sqrt{3}=\sqrt{3}+\sqrt{3}\)
ta có \(\sqrt{5}>\sqrt{3}\)và\(\sqrt{7}>\sqrt{3}\)=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
a: \(\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\)
\(\left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
mà \(-2\sqrt{105}>-2\sqrt{120}\)
nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b: \(\left(\sqrt{2}+\sqrt{8}\right)^2=10+2\cdot4=16=12+4\)
\(\left(3+\sqrt{3}\right)^2=12+6\sqrt{3}\)
mà \(4< 6\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{8}< 3+\sqrt{3}\)
a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)
=> A2=8+2\(\sqrt{3}\)
B=\(\sqrt{3}+1\)=> B2=10+2\(\sqrt{3}\)
=>A>B
a) Bình phương lên,ta so sánh \(\left(\sqrt{5}+\sqrt{7}\right)^2=5+2\sqrt{35}+7\text{ và }12\)
Xét hiệu hai vế \(\left(\sqrt{5}+\sqrt{7}\right)^2-12=2\sqrt{35}>0\) nên ....
b) \(14=\sqrt{14^2}=\sqrt{196}>\sqrt{195}=\sqrt{13}.\sqrt{15}\)
c) \(\left(\sqrt{8}+3\right)^2=8+2.\sqrt{72}+9;\left(6+\sqrt{2}\right)^2=36+2\sqrt{72}+2\)
\(\left(8+\sqrt{3}\right)^2-\left(6+\sqrt{2}\right)^2=\left(8+9\right)-\left(36+2\right)< 0\)
Do đó \(\left(8+\sqrt{3}\right)^2< \left(6+\sqrt{2}\right)^2\) suy ra \(\left(8+\sqrt{3}\right)< \left(6+\sqrt{2}\right)\)
d) So sánh \(\sqrt{27}+\sqrt{6}\text{ và }\sqrt{48}-1\)
Dễ chứng minh \(\sqrt{27}+\sqrt{6}> \sqrt{48}-1\)
Suy ra \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\) (thêm 1 vào mỗi vế)
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: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
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~