So sánh
1-\(\frac{1}{\sqrt{8}}\)và \(\frac{3}{4}\)
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1) `-3\sqrt13=-3\sqrt13`
`-9=-3\sqrt9`
`\sqrt13>\sqrt9`
`=> -3\sqrt13 < -3\sqrt9`
`=> -3\sqrt13 < 9`.
2) `\sqrt15 < \sqrt16`
`<=> \sqrt15-1 < \sqrt16-1`
`<=> \sqrt15-1 < 3 < \sqrt10`
`=> \sqrt15-1 <\sqrt10`
3) `5=4+1=\sqrt16+1`
`\sqrt8+1=\sqrt8+1`
`=> 5>\sqrt8+1`
1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)
3) Ta có: \(5^2=25=9+16\)
\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)
mà \(16>4\sqrt{2}\)
nên \(5>2\sqrt{2}+1\)
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: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
Lời giải:
Xét số hạng tổng quát của tổng trên:
\(\frac{n+(n+2)+\sqrt{n(n+2)}}{\sqrt{n}+\sqrt{n+2}}=\frac{(2n+2+\sqrt{n(n+2)})(\sqrt{n+2}-\sqrt{n})}{(\sqrt{n}+\sqrt{n+2})(\sqrt{n+2}-\sqrt{n})}\)
\(=\frac{(n+2)\sqrt{n+2}-n\sqrt{n}}{2}\)
Áp dụng vào bài:
\(P=\frac{3\sqrt{3}-1}{2}+\frac{5\sqrt{5}-3\sqrt{3}}{2}+\frac{7\sqrt{7}-5\sqrt{5}}{2}+...+\frac{121\sqrt{121}-119\sqrt{119}}{2}\)
\(=\frac{121\sqrt{121}-1}{2}=665\)
1) \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)
\(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0
=> A=3
2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)
\(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)
Mà A >0
=> A=2
Mà 4>3
=> \(\sqrt{4}=2>\sqrt{3}\)
=> \(A>\sqrt{3}\)
Ta có : \(\frac{3}{\sqrt{6}-\sqrt{3}}=\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{\left(\sqrt{6}-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{3}\right)}=\sqrt{6}+\sqrt{3}\)
và \(\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}=\sqrt{7}-\sqrt{3}\)
\(\Rightarrow\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\sqrt{6}+\sqrt{7}=\frac{\left(\sqrt{6}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\)
Vậy hai giá trị trên bằng nhau.
ta có +,\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{3}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}\)\(=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)
+,\(\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1\left(\sqrt{7}+\sqrt{6}\right)}{1}=\sqrt{7}+\sqrt{6}\)
vậy \(\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
Ta có: \(\frac{1}{8}>\frac{1}{9}\) => \(\sqrt{\frac{1}{8}}>\sqrt{\frac{1}{9}}\)hay \(\frac{1}{\sqrt{8}}>\frac{1}{\sqrt{9}}=\frac{1}{3}\)
=> \(1-\frac{1}{\sqrt{8}}< 1-\frac{1}{3}\)
\(\frac{3}{4}=1-\frac{1}{4}\)
Do \(\frac{1}{3}>\frac{1}{4}\) => \(1-\frac{1}{3}< 1-\frac{1}{4}\)
hay \(1-\frac{1}{\sqrt{8}}< \frac{3}{4}\)
Bài làm:
Ta có: \(1-\frac{1}{\sqrt{8}}< 1-\frac{1}{\sqrt{9}}=1-\frac{1}{3}< 1-\frac{1}{4}=\frac{3}{4}\)
\(\Rightarrow1-\frac{1}{\sqrt{8}}< \frac{3}{4}\)