CMR A>1 với
A=\(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+...+\frac{1}{\sqrt{1999.1}}\)
theo cosi đc ko?
Cho A=
\(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+\frac{1}{\sqrt{3.1997}}+\dots+\frac{1}{\sqrt{1999.1}}\)
Hãy so sánh A và 1,999
Áp dụng bất đẳng thức Cô-si:
\(\frac{1}{\sqrt{1\cdot1999}}\ge\frac{1}{\frac{1+1999}{2}}=\frac{1}{1000}\)
Vì dấu "=" không xảy ra nên \(\frac{1}{\sqrt{1\cdot1999}}>\frac{1}{1000}\)
Tương tự ta có : \(\frac{1}{\sqrt{2\cdot1998}}>\frac{1}{1000};...;\frac{1}{\sqrt{1999\cdot1}}>\frac{1}{1000}\)
\(\Rightarrow\frac{1}{\sqrt{1\cdot1999}}+\frac{1}{\sqrt{2\cdot1998}}+...+\frac{1}{\sqrt{1999\cdot1}}>\frac{2000}{1000}=2>1,999\)
Vậy...
CMR A=\(\dfrac{1}{\sqrt{1.1999}}+\dfrac{1}{\sqrt{2.1998}}+....+\dfrac{1}{\sqrt{1999.1}}>1,999\)
\(A=\dfrac{1}{\sqrt{1.1999}}+\dfrac{1}{\sqrt{2.1998}}+...+\dfrac{1}{\sqrt{1999.1}}>\dfrac{1}{\dfrac{1+1999}{2}}+\dfrac{1}{\dfrac{2+1998}{2}}+...+\dfrac{1}{\dfrac{1999+1}{2}}\)
\(=\dfrac{1}{1000}+\dfrac{1}{1000}+...+\dfrac{1}{1000}=1,999\)
1) CMR \(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+\frac{1}{\sqrt{3.1997}}+...+\frac{1}{\sqrt{1999.1}}\ge1,999\)
2) CMR \(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{95\sqrt{94}+94\sqrt{95}}< 1\)
3) CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
4) CMR \(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
cho A= (1/căn 1.1999)+(1/2.1998)+...+(1/1999.1)
chưng minh A>1999
rút gọn biểu thức
\(\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}vớia>0;a\ne1\)
\(\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=\left(\frac{0}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0\)
\(A=\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) đkxđ:\(a>0;a\ne1\)
\(A=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}}\)\
\(A=0\)
Rút gọn rồi tính giá trị của biểu thức
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}\div\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}vớia=7,25;b=3,25\)
\(\frac{a-b}{\sqrt{a\times\left(a+2\times b\right)+b^2}}\div\sqrt{\frac{\left(a-b\right)^2}{a\times\left(a+b\right)}}vớia>b>0và\frac{a}{b}=\frac{9}{7}\)
\(\frac{x-1}{\sqrt{y}-1}\times\sqrt{\frac{\left(y-2\times\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}vớix=\frac{-1}{2};y=121\); giúp mk vs
RÚT GỌN BIỂU THỨC SAU
\(A=\frac{\sqrt{a}+a\sqrt{a}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(vớia\ge0,b\ge0;ab\ne1\right)\)
\(B=\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)
Rút gọn biểu thức
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0; a ≠ b)
b) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)vớia,b>0\)
Phá ngoặc được \(T=2+\frac{1}{a}+\frac{1}{b}+a+b+\frac{a}{b}+\frac{b}{a}=2+\frac{a+b}{ab}+a+b+\frac{a}{b}+\frac{b}{a}\)
Theo bdt cosi ta có \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow T\ge4+\frac{a+b}{ab}+a+b\)
Ta có \(\frac{a+b}{ab}+a+b=\frac{a+b}{2ab}+\left(a+b\right)+\frac{a+b}{2ab}\) Theo bdt cosi
\(\frac{a+b}{2ab}+\left(a+b\right)\ge2\sqrt{\frac{\left(a+b\right)^2}{2ab}}\ge2\sqrt{\frac{4ab}{2ab}}=2\sqrt{2}\)
Lại có \(1=a^2+b^2\ge2ab\Rightarrow\frac{1}{ab}\ge2\Rightarrow\frac{1}{\sqrt{ab}}\ge\sqrt{2}\)
\(\frac{a+b}{2ab}\ge\frac{2\sqrt{ab}}{2ab}=\frac{1}{\sqrt{ab}}\ge\sqrt{2}\) \(\Rightarrow T\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{\sqrt{2}}\)