\(\left(-3\right)^2.\frac{1}{3}-\sqrt{49}+\left(-5\right)^2:\sqrt{25}\)
dành cho đứa học bình thường
PD
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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
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+....+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
Cho M= \(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}\)+........+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)
Cm :M<\(\frac{2}{5}\)
CM;
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+....+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
\(\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}\)+\(\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}\)+...+\(\frac{1}{49\left(\sqrt{24}+\sqrt{25}\right)}\)<\(\frac{2}{5}\)
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)^2cdotfrac{4}{11}+frac{7}{11}cdotleft(-frac{1}{3}right)^2
c, left(-frac{1}{7}right)^0-2frac{4}{9}cdotleft(frac{2}{3}right)^2
d,frac{2^7cdot9^2}{3^3cdot2^5}
e,left(frac{1}{3}-frac{5}{6}right)^2+frac{5}{6}:2
f,left(9frac{2}{4}:5,2+3.4cdot2frac{7}{34}right):left(-1frac{9}{16}right)
g,sqrt{25}-3sqrt{frac{4}{9}}
h,left(-2right)^2+sqrt{36}-sqrt{9}+sqrt...
Đọc tiếp
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,\(\frac{\sqrt{3^2+\sqrt{39^2}}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
Chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \frac{3}{7}\)
Tính
a) \(2\sqrt{\frac{25}{16}}-3\sqrt{\frac{49}{36}}+4\sqrt{\frac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\frac{1}{2}}\right)^2+\frac{1}{16}.\left(\sqrt{\frac{3}{4}}\right)^2\)
c) \(\frac{2}{3}\sqrt{\frac{81}{16}}-\frac{3}{4}\sqrt{\frac{64}{9}}+\frac{7}{5}.\sqrt{\frac{25}{196}}\)
a) = \(\frac{7}{2}\)
b) = \(\frac{643}{64}\)
c) = 0
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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 <\(\frac{3}{7}\)
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