1) TÌM GTNN CỦA:
a) ( x - 5 )\(^4\)+ \(\frac{14}{17}\)
b) ( \(\frac{3}{7}\)- 14x )\(^2\)- \(\frac{214}{979}\)
2) SO SÁNH
\(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) và 56
So sánh \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) và 60
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)
\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)
Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) < 60 nha.
Bài 1: So sánh a) 31^60 và 17^74
b) \(\sqrt{12}\)+\(\sqrt{20}\)+\(\sqrt{30}\)+\(\sqrt{42}\)+\(\sqrt{56}\)+\(\sqrt{72}\)+\(\sqrt{90}\)\(\sqrt{110}\)và 56
Bài 2: Cho tỉ lệ thức: \(\frac{\overline{ab}}{\overline{bc}}=\frac{a}{c}\)
CMR: \(\frac{\overline{abbb...b}}{\overline{bbb...bc}}=\frac{a}{c}\)(có n chữ số b; n là 1 số tự nhiên)
các bạn trình bày cách làm giùm mình với
So sánh
a) \(\sqrt{37}+\sqrt{83}\) và 15
b) \(\sqrt{48}+\sqrt{80}\) và 16
c) \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) và 56
Không dùng máy tính hãy cm
A= \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}< 60\)
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
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)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)
Chứng minh rằng:
a)\(\sqrt{1}+\sqrt{2}+...+\sqrt{8}< 24\)
b)\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10\)
c)\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}< 30\)
b, \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
.............................................
Cộng với vế 99 của BĐT trên, ta được:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99.\frac{1}{10}=\frac{99}{10}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}=\frac{1}{10}=\frac{100}{10}=10\)
Wrecking Ball đã làm đúng
to ra kết quả giống cậu : Wrecking Ball
là đáp án đúng
tk nha ( chúc bn học gioi )
Rút gọn biểu thức
1)\(\frac{15}{3\sqrt{20}}\)
2) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}\)
3) \(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
4) \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
5) \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)
6)\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
7) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
8)\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
9) \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
10) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
11) \(\sqrt{\sqrt{10}+1}.\sqrt{\sqrt{10}-1}\)
12) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
13) \(\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{12}}\)
14) \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\sqrt{6}\)
15 ) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
16) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
17) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
Tính:
a/ \(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{12}{3-\sqrt{3}}\)
b/ \(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{7}+\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
c/ \(\frac{\sqrt{14}-\sqrt{17}}{1-\sqrt{2}}\)
d/ \(\frac{3\sqrt{2}-3}{\sqrt{2}-1}\)
e/ \(\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\)