Chứng minh:
\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+.....+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
chứng minh bất đẳng thức:
\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{9}+\sqrt{11}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)<\(\dfrac{9}{4}\)
CMR:\(\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{9}+\sqrt{11}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
chứng minh rằng:\(\dfrac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\dfrac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\dfrac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \dfrac{3}{7}\)
Bạn tham khảo câu số 9:
mọi người giúp em mấy bài này với ạ =((( - Hoc24
Tính giá trị biểu thức:
A= \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
Nhận xét 1: từng hạng tử của A có dạng:
\(\dfrac{1}{\sqrt{x}+\sqrt{x+2}}\left(x\ge3\right)\)
Nhận xét 2:
\(\left(\sqrt{x+2}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{x+2}\right)=\left(x+2\right)-x=2\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+\sqrt[]{x+2}}=-\sqrt{x}+\sqrt{x+2}\)
Áp dụng vào A:
\(2A=\dfrac{2}{\sqrt{3}+\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{7}}+...+\dfrac{2}{\sqrt{97}+\sqrt{99}}\)
\(=\left(-\sqrt{3}+\sqrt{5}\right)+\left(-\sqrt{5}+\sqrt{7}\right)+...+\left(-\sqrt{97}+\sqrt{99}\right)\)
\(=-\sqrt{3}+\sqrt{99}\Leftrightarrow A=-2\sqrt{3}+2\sqrt{99}\)
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=
\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)}+\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\cdot\left(\sqrt{9}-\sqrt{7}\right)}+...+\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{97}+\sqrt{99}\right)\cdot\left(\sqrt{99}-\sqrt{97}\right)}\)
= \(\dfrac{\sqrt{5}-\sqrt{3}}{5-3}+\dfrac{\sqrt{7}-\sqrt{5}}{7-5}+\dfrac{\sqrt{9}-\sqrt{7}}{9-7}+...+\dfrac{\sqrt{99}-\sqrt{97}}{99-97}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+\dfrac{\sqrt{9}-\sqrt{7}}{2}+...+\dfrac{\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{1}{2}\cdot\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\cdot\left(-\sqrt{3}+\sqrt{99}\right)\)
= \(\dfrac{3\sqrt{11}-\sqrt{3}}{2}\)
đó là ý kiến của mik còn đúng hay sai là tùy bn
Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
Rút gọn:
a) \(A=\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+... +\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2006\sqrt{2005}+2005\sqrt{2006}}+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
Tiếp phần b ( do máy lag) :3
Cộng 2 vế với nhau, ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)
a) A=\(\dfrac{1}{\sqrt{3}+\sqrt{5}}\)+\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)+\(\dfrac{1}{\sqrt{7}+\sqrt{9}}\)+...+\(\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)+\(\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)+\(\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\left(\sqrt{9}-\sqrt{7}\right)}\)+...+\(\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
vậy A=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
Tính giá trị của các biểu thức:
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
= \(\dfrac{1}{2}\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\left(\sqrt{99}-\sqrt{3}\right)\)
B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35
= 33 + 2 + 333 + 2 + 3333 + 2 + ... + 333...33 + 2
= 2 . 99 + (33 + 333 + 3333 + ... + 333...3)
= 198 + \(\dfrac{1}{3}\)(99 + 999 + 9999 + ... + 999...99)
= 198 + \(\dfrac{1}{3}\)(102 - 1 + 103 - 1 + 104 - 1 + ... + 10100 - 1)
= \(\left(\dfrac{10^{101}-10^2}{27}\right)+165\)
Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)