Tính giá trị gần đúng đến 5 chữ số thập phân của
M = \(\frac{\sqrt{3}}{\sqrt[3]{2}+\sqrt[3]{4}}+\frac{\sqrt{5}}{\sqrt[3]{4}+\sqrt[3]{6}}+....+\frac{\sqrt{59}}{\sqrt[3]{58}+\sqrt[3]{60}}\)
tính tổng:
\(\frac{\sqrt{3}}{\sqrt[3]{2}+\sqrt[3]{4}}+\frac{\sqrt{5}}{\sqrt[3]{4}+\sqrt[3]{6}}+\frac{\sqrt{7}}{\sqrt[3]{6}+\sqrt[3]{8}}+...+\frac{\sqrt{57}}{\sqrt[3]{56}+\sqrt[3]{58}}+\frac{\sqrt{59}}{\sqrt[3]{58}+\sqrt[3]{60}}\)
Bài 1: Tính giá trị của biểu thức:\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyên
A = \(\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
B = \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Tính giá trị của biểu thức:
Q = \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{3}+\sqrt{4}}{1+\sqrt{3}+\sqrt{4}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2012}+\sqrt{2013}}{1+\sqrt{2012}+\sqrt{2013}}\)
!@#$%^&*()_+\ [];'{}
đầu hàng tại chỗ !
hiiiii
NX \(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}\) =\(\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}-1\right)}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}+1\right)^2}\)
=\(\frac{\left(\left(\sqrt{n+1}-\sqrt{n}\right)^2-1^2\right)}{n+1-n-1-2\sqrt{n}}\) \(=\frac{n+1+n-2\sqrt{\left(n+1\right)n}-1}{-2\sqrt{n}}=\frac{2n-2\sqrt{n\left(n+1\right)}}{-2\sqrt{n}}\)
=\(\frac{n-\sqrt{n\left(n+1\right)}}{-\sqrt{n}}=\frac{n}{-\sqrt{n}}+\frac{\sqrt{n\left(n+1\right)}}{\sqrt{n}}=-\sqrt{n}+\sqrt{n+1}\)
thay vao Q ta co
Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)
Tính giá trị gần đúng đến chữ số chữ số ở phần thập phân:
\(A=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\sqrt[5]{5+\sqrt[6]{6+\sqrt[7]{7+\sqrt[8]{8+\sqrt[9]{9}}}}}}}}\)
Các bạn làm ơn trình bày giống như trình bày toán casio nhé mình chỉ mới học thôi chưa hết cách trình bày 1 bay giải casio như thế nào cả
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\(y=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}-\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{4}}}\)
Mọi người giải hộ mình nhé. Tính giá trị biểu thức trên
Bài 1: Tính giá trị của biểu thức:
1)\(H=\sqrt[3]{3+\sqrt{9+\frac{125}{7}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{7}}}\)
2)\(P=\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}\)
Bài 2: Tính giá trị biểu thức: \(Q=\sqrt[10]{\frac{19+6\sqrt{10}}{2}}.\sqrt[5]{3\sqrt{2}-2\sqrt{5}}\)
\(K=\frac{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}-\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}}\)
\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)
Tính \(\frac{2\sqrt{3}-4}{\sqrt{3}-1}+\frac{2\sqrt{2}-1}{\sqrt{2}-1}-\frac{1+\sqrt{6}}{\sqrt{2}+3}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)