CMR A=\(\dfrac{1}{\sqrt{1.1999}}+\dfrac{1}{\sqrt{2.1998}}+....+\dfrac{1}{\sqrt{1999.1}}>1,999\)
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...
a, Cho S=\(\dfrac{1}{\sqrt{1.1998}}+\dfrac{1}{\sqrt{2.1997}}+...+\dfrac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\dfrac{1}{\sqrt{198-1}}\). Hãy so sánh S và 2\(\dfrac{1998}{1999}\)
b, Cho A=\(\dfrac{1}{\sqrt{1.1999}}+\dfrac{1}{\sqrt{2.1998}}+\dfrac{1}{\sqrt{3.1997}}+...+\dfrac{1}{\sqrt{199-1}}\). Hãy so sánh A với 1,999
Câu a :
Áp dụng BĐT \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\left(a\ne b;a,b>0\right)\) ta có :
\(\dfrac{1}{\sqrt{1.1998}}>\dfrac{2}{1+1998}=\dfrac{2}{1999}\)
\(\dfrac{1}{\sqrt{2.1997}}>\dfrac{2}{2+1997}=\dfrac{2}{19999}\)
.......................................................
\(\dfrac{1}{\sqrt{1998.1}}>\dfrac{2}{1998+1}=\dfrac{2}{1999}\)
Cộng tất cả vế với nhau ta được : \(P>2.\dfrac{1998}{1999}\)
\(\Rightarrowđpcm\)
Câu a, b sao tính chất cái cuối khác những cái còn lại thế. Vậy sao biết tới đâu thì nó dừng.
CMR A>1 với
A=\(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+...+\frac{1}{\sqrt{1999.1}}\)
theo cosi đc ko?
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
CMR:
\(\sqrt{a}-\dfrac{a-\dfrac{1}{a^2}}{\sqrt{a}-\dfrac{1}{\sqrt{a}}}+\dfrac{1-\dfrac{1}{a^2}}{\sqrt{a}+\dfrac{1}{\sqrt{a}}}+\dfrac{2}{a\sqrt{a}}=0\)
4.
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)-\left(\dfrac{1}{x+\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
a. Rút gọn A.
b. Tính x khi \(A=\dfrac{1}{2}\)
5. CMR
\(\left(\dfrac{\sqrt{30}}{\sqrt{3}}-\dfrac{\sqrt{20}}{\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right).\sqrt{4+\sqrt{15}}=2\)
nhanh lên nha
Cho biểu thức A:
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+1+\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2\sqrt{x}}\)
a) Rút gọn A.
b) cmr: \(A< \dfrac{2}{3}\)
a) ĐKXĐ: \(x>0,x\ne1\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}=\dfrac{2\sqrt{x}}{x+\sqrt{x}+1}\)
CMR \(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\ \dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\ .........\\ \dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+....+\dfrac{1}{\sqrt{100}}\)( 100 phân số \(\dfrac{1}{\sqrt{100}}\) )
hay \(A>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+....+\dfrac{1}{10}\)(100 phân số \(\dfrac{1}{10}\) )
\(\Rightarrow A>\dfrac{100}{10}\\ \Rightarrow A>10\)
KL : Vậy ....