\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+...+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
Rút gọn :
M = \(\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}+....+\frac{1}{49.\left(\sqrt{24}+\sqrt{25}\right)}\)
Chứng minh
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+....+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
CM;
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+....+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
Chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \frac{3}{7}\)
Bài 1 : Cho \(S=\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}\)
So sánh S với \(\frac{3}{7}\)
\(tacó:...\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}>\frac{1}{3.2}=\frac{1}{\left(1+2.1\right).2.1}\)
\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}>\frac{1}{5.4}=\frac{1}{\left(1+2.2\right).2.2}\)
\(\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}>\frac{1}{7.6}=\frac{1}{\left(1+2..3\right).2.3}\)
....
\(\frac{1}{49.\left(\sqrt{48}+\sqrt{49}\right)}>\frac{1}{49.48}=\frac{1}{\left(1+2.48\right).2.48}\)
cộng vế theo vế ta đươc S =\(\frac{1}{\left(1+2.1\right).2}+\frac{1}{\left(1+2.2\right).2.2}+...+\frac{1}{\left(1+2.48\right).48.2}\)
\(=\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{10}+\frac{1}{21}+\frac{1}{36}+...+\frac{1}{4656}\right)\) < \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4656}\right)\)
mà lại có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+..+\frac{1}{4656}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9312}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{96.97}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{97}=\frac{1}{2}-\frac{1}{97}=\frac{95}{194}\)
vậy S < \(\frac{95}{194}\)
mà \(\frac{95}{194}< \frac{3}{7}\)
=> S < \(\frac{3}{7}\)
KẾT LUẬN : S <\(\frac{3}{7}\)
Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
Sao làm hổng ai bảo đú.n/g vậy :(((
Bài 1: Thực hiện phép tính:
a,\(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{7}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
b,\(\left(-\frac{1}{3}\right)^2\cdot\frac{4}{11}+\frac{7}{11}\cdot\left(-\frac{1}{3}\right)^2\)
c, \(\left(-\frac{1}{7}\right)^0-2\frac{4}{9}\cdot\left(\frac{2}{3}\right)^2\)
d,\(\frac{2^7\cdot9^2}{3^3\cdot2^5}\)
e,\(\left(\frac{1}{3}-\frac{5}{6}\right)^2+\frac{5}{6}:2\)
f,\(\left(9\frac{2}{4}:5,2+3.4\cdot2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)\)
g,\(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
h,\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
i,\(\left(-\frac{1}{2}\right)^4+\left|-\frac{2}{3}\right|-2007^0\)
k,\(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
m,\(\left(-3\right)^2\cdot\frac{1}{3}-\sqrt{49}+\left(-5\right)^3:\sqrt{25}\)
n,\(\frac{\sqrt{3^2+\sqrt{39^2}}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
Rút Gọn
a,\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
b,\(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
c,\(\left(\sqrt{12}+2\sqrt{27}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
d,\(\left(\sqrt{18}+\sqrt{0,5}-3\sqrt{\frac{1}{3}}\right)-\left(\sqrt{\frac{1}{8}-\sqrt{75}}\right)\)
e,\(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
f,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
g,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
h,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
i,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
j,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+7\right):\sqrt{7}\)
Tính nhanh:
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{25}\right).\left(-\frac{4}{15}\right)}{\:\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\left(\frac{5}{7}\right)}\)