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\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)
Lời giải :
Xét dạng tổng quát sau :
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Từ đó ta có hướng giải quyết bài toán :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}\)
\(A=\sqrt{25}-\sqrt{1}\)
\(A=4\)
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
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: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}=\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)\(=1-\frac{1}{5}=\frac{4}{5}\)
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:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
Sửa đề:
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}=\sqrt{25}-1=4\)