Rút gọn:
A = \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
B = \(\frac{1}{\sqrt{1}-\sqrt{2}}+\frac{1}{\sqrt{2}-\sqrt{3}}+...+\frac{1}{\sqrt{24}-\sqrt{25}}\)
Bt: Tính
a) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
b) C/m: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
c) C/m: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}
a, bạn chỉ cần lập công thức tông quát :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cái này bạn chỉ cần trục căn thức ở mẫu chưng minh xong áp dụng vào luôn là ra
a, kq : 4/5
b,\(1-\frac{1}{\sqrt{n+1}}\)
c,d chưa nghĩ ra
Bt: Tính
a) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
b) C/m: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
c) C/m: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}<2\)
d) C/m: \(\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}\)
ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{\left(n+1\right)n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{\left(n+1\right)n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
nên: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}=\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)\(=1-\frac{1}{5}=\frac{4}{5}\)
CMR
\(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+........+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
\(\sqrt{1}+\sqrt{2}+.......+\sqrt{n}< n\sqrt{\frac{n+1}{2}}\)
\(\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+.........+\frac{\sqrt{25}-\sqrt{24}}{24+25}< \frac{2}{5}\)
a. Chứng minh : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. Áp dụng : Tính giá trị của biểu thức :
\(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
cảm ơn các bạn trước nhé!
a) CMR: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\) với \(n\in N\)*
b) tính \(B=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+......+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
a/ Quy đồng vế phải, hình như lộn mẫu cuối là căn 2 của (n+1) mới đúng
\(VP=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}}{\sqrt{n+1}.\sqrt{n}}\)
\(=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right).\sqrt{n+1}.\sqrt{n}}=\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=VT\)
\(B=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=1-\frac{1}{\sqrt{25}}\)
Rút gọn các biểu thức,
a> A= \(\frac{1}{\sqrt{1}+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\)+ \(\frac{1}{\sqrt{3}+\sqrt{4}}\)+ ......... + \(\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
b> B= \(\frac{1}{\sqrt{1}-\sqrt{2}}\)- \(\frac{1}{\sqrt{2}-\sqrt{3}}\)- \(\frac{1}{\sqrt{3}-\sqrt{4}}\)- .......... - \(\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+.......+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n}-1\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+........+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(=\sqrt{2}-\sqrt{1}+...........+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
bài B tương tự
Help me pleases ,thanks before
1) c/m \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) với mọi số nguyên dương n
2)cho A=\(\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+....+\frac{\sqrt{25}-\sqrt{24}}{24+25}\)
C/m \(A< \frac{2}{5}\)
3)Cho 3 số a,b,c dương,c/m
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)
Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
Bài 2:
Ta có:
\(\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Nên:
\(A< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\right)=\frac{1}{2}\left(1-\frac{1}{5}\right)=\frac{2}{5}\)
rút gọn:\(\frac{1}{\sqrt{1}-\sqrt{2}}+\frac{1}{\sqrt{2}-\sqrt{3}}+...+\frac{1}{\sqrt{24}-\sqrt{25}}\)
Nhân liên hiệp ta được :
\(\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}+\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{\sqrt{24}+\sqrt{25}}{\left(\sqrt{24}-\sqrt{25}\right)\left(\sqrt{24}+\sqrt{25}\right)}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{1-2}+\frac{\sqrt{2}+\sqrt{3}}{2-3}+...+\frac{\sqrt{24}+\sqrt{25}}{24-25}\)
\(=-\sqrt{1}-\sqrt{2}-\sqrt{2}-\sqrt{3}-....-\sqrt{24}-\sqrt{25}\)
\(=-\left[\frac{\left(\sqrt{25}+\sqrt{1}\right).25}{2}+\frac{\left(\sqrt{24}+\sqrt{2}\right).23}{2}\right]\)
\(=...\)
Rút gọn : \(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}-...-\frac{1}{\sqrt{24}-\sqrt{25}}\)