chứng minh B=\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{78}+\sqrt{79}}>4\)
Chứng minh rằng: \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}>4\)
Đặt \(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
Ta có: \(\frac{1}{1+\sqrt{2}}>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)
...
\(\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
Cộng các bất đẳng thức trên lại với nhau, ta được:
\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{81}-1\right)=\frac{1}{2}\cdot\left(9-1\right)=\frac{1}{2}\cdot8=4\)
\(\Leftrightarrow A>4\)(đpcm)
Chứng minh:
B=\(\frac{1}{\sqrt{1}+\sqrt{2}}\) + \(\frac{1}{\sqrt{3}+\sqrt{4}}\) +\(\frac{1}{\sqrt{5}+\sqrt{6}}\) +.............+\(\frac{1}{\sqrt{77}+\sqrt{78}}\) +\(\frac{1}{\sqrt{79}+\sqrt{80}}\) >4
Giúp mk nha.....!!!!!!
Chứng minh rằng :
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{78}+\sqrt{80}}>4\)
1. Chứng minh rằng
\(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}>4\)
2. Chứng minh rằng
\(\frac{\sqrt{1}}{1}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}+...+\frac{\sqrt{200}}{200}>10+5\sqrt{2}\)
3. Cho a >= 1, b >= 1, chứng minh rằng
\(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
4. Giải phương trình
\(\sqrt{\left(x^2-2x+5\right)\left(x^2-4x\right)+7}+x^2-3x+6\)
LÀM PHIỀN M.N GIÚP MK. XIN CẢM ƠN !!!
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
CMR \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}>4\)
Trước hết , ta cần chứng minh \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*) (Bạn tự chứng minh)
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(\Rightarrow2A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
Áp dụng (*) :\(\Rightarrow2A>\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{4}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)+\left(\sqrt{81}-\sqrt{80}\right)\)
\(\Rightarrow2A>\sqrt{81}-1=8\Rightarrow A>4\)(đpcm)
CMR:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+....+\frac{1}{\sqrt{79}+\sqrt{80}}>4\)
\(\frac{1}{\sqrt{1}+\sqrt{2}}+....\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\) (40 số)
................................................................\(>\frac{40}{10}=4\)
=>đpcm
hc tốt
ko chắc lắm :)
cho A = \(^{\frac{1}{\sqrt{1+\sqrt{2}}}+\frac{1}{\sqrt{3+\sqrt{4}}}+...+\frac{1}{\sqrt{79+\sqrt{80}}}}\)
chứng minh A>4
dùng cách trục căn thức là ra
Chứng tỏ P < \(\frac{8}{9}\)
P=\(\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{4}+\frac{\sqrt{7}-\sqrt{5}}{6}+...+\frac{\sqrt{81}-\sqrt{79}}{80}\)
1.chứng minh các đẳng thức sau:
a.\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}\)
b.\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=-2\)
a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)
--> dpcm
b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)
=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(7-5\right)\)
=-1.2
=-2