chứng minh rằng
a) \(\sqrt{4+\sqrt{4+....+\sqrt{4}+\sqrt{ }4}}< 3\)
b)\(\sqrt{1^3+2^3+...+n^3}=1+2+...+n\)
Những câu hỏi liên quan
chứng minh rằng với số tự nhiên n,n lớn hơn 4 ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
Đúng 2
Bình luận (0)
Câu 4:
a. Chứng minh rằng: \(\sqrt{22-12\sqrt{2}}\) + \(\sqrt{6+4\sqrt{2}}\) = 4\(\sqrt{2}\)
b. Chứng minh rằng: \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}\) = \(\sqrt{n+1}\) - \(\sqrt{n}\)
\(a,\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\\ =3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\\ b,\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}\\ =\dfrac{\sqrt{n}-\sqrt{n+1}}{-1}=\sqrt{n+1}-\sqrt{n}\)
Đúng 0
Bình luận (0)
a) \(\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\)
b) \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Đúng 1
Bình luận (0)
Cho a,b,c>0 thỏa mãn abc=1. Chứng minh rằng
\(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\)
Cho các số thực dương a,b,c thỏa mãn abc=1.Chứng minh rằng:
\(\frac{1}{\sqrt{a^4-a^3+ab-2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4+c^3+ac+2}}\le\sqrt{3}\)
Đề: \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\) ???
*Ta chứng minh : \(x^4-x^3+2\ge x+1\forall x>0\)
\(\Leftrightarrow x^4-x^3-x+1\ge0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\ge0\) ( đúng )
Do đó: \(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\) \(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=1\)
Đúng 0
Bình luận (0)
1. Tính giá trị biểu thức: Asqrt{a^2+4ab^2+4b}-sqrt{4a^2-12ab^2+9b^4} với asqrt{2} ; b1
2. Đặt Msqrt{57+40sqrt{2}} ; Nsqrt{57-40sqrt{2}}. Tính giá trị của các biểu thức sau:
a) M-N
b) M^3-N^3
3. Chứng minh: left(frac{xsqrt{x}+3sqrt{3}}{x-sqrt{3x}+3}-2sqrt{x}right)left(frac{sqrt{x}+sqrt{3}}{3-x}right)1 (với xge0 và xne3)
4. Chứng minh: frac{left(sqrt{a}-sqrt{b}right)^2+4sqrt{ab}}{sqrt{a}+sqrt{b}}.frac{asqrt{b}-bsqrt{a}}{sqrt{ab}}a-b (a 0 ; b 0)
5. Chứng minh: sqrt{9+4sqrt{2}}2sqrt{2}+1 ;...
Đọc tiếp
1. Tính giá trị biểu thức: \(A=\sqrt{a^2+4ab^2+4b}-\sqrt{4a^2-12ab^2+9b^4}\) với \(a=\sqrt{2}\) ; \(b=1\)
2. Đặt \(M=\sqrt{57+40\sqrt{2}}\) ; \(N=\sqrt{57-40\sqrt{2}}\). Tính giá trị của các biểu thức sau:
a) M-N
b) \(M^3-N^3\)
3. Chứng minh: \(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)=1\) (với \(x\ge0\) và \(x\ne3\))
4. Chứng minh: \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=a-b\) (a > 0 ; b > 0)
5. Chứng minh: \(\sqrt{9+4\sqrt{2}}=2\sqrt{2}+1\) ; \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}\) ; \(3-2\sqrt{2}=\left(1-\sqrt{2}\right)^2\)
6. Chứng minh: \(\left(\frac{1}{2\sqrt{2}-\sqrt{7}}-\left(3\sqrt{2}+\sqrt{17}\right)\right)^2=\left(\frac{1}{2\sqrt{2}-\sqrt{17}}-\left(2\sqrt{2}-\sqrt{17}\right)\right)^2\)
7. Chứng minh đẳng thức: \(\left(\frac{3\sqrt{2}-\sqrt{6}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}=-\frac{4}{3}\)
8.Chứng minh: \(\frac{2002}{\sqrt{2003}}+\frac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
9. Chứng minh rằng: \(\sqrt{2000}-2\sqrt{2001}+\sqrt{2002}< 0\)
10. \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) ; \(\frac{7}{5}< \frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}< \frac{29}{30}\)
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
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}......+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
Xem thêm câu trả lời
Chứng minh rằng
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n}+1\)
cảm ơn
Lời giải:
Sửa đề: CMR:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n}-1\)
-------------------------------
Sử dụng PP liên hợp ta có:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+....+\frac{\sqrt{n}-\sqrt{n-1}}{(\sqrt{n-1}+\sqrt{n})(\sqrt{n}-\sqrt{n-1})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+....+\frac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
Ta có đpcm.
Đúng 0
Bình luận (0)
1) Chứng minh rằng: 1+dfrac{1}{2sqrt{2}}+dfrac{1}{3sqrt{3}}+...+dfrac{1}{nsqrt{n}} 2sqrt{2}left(nin Nright)
2) Chứng minh rằng: dfrac{2}{3}+sqrt{n+1} 1+sqrt{2}+sqrt{3}+...+sqrt{n} dfrac{2}{3}left(n+1right)sqrt{n}
3) 2sqrt{n}-3 dfrac{1}{sqrt{2}}+dfrac{1}{sqrt{3}}+...+dfrac{1}{sqrt{n}} 2sqrt{n}-2
4) dfrac{sqrt{2}-sqrt{1}}{2+1}+dfrac{sqrt{3}-sqrt{2}}{3+2}+...+dfrac{sqrt{n+1}-sqrt{n}}{n+1+n} dfrac{1}{2}left(1-dfrac{1}{sqrt{n+1}}right)
Đọc tiếp
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Chứng minh rằng với mọi số nguyên dương n ta đều có \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+\dfrac{1}{5\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(n+1)\sqrt{n}}<\frac{(\sqrt{n+1}-\sqrt{n}).2\sqrt{n+1}}{(n+1)\sqrt{n}}\)
Hay \(\frac{1}{(n+1)\sqrt{n}}< \frac{2\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài toán:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}+....+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}=2-\frac{2}{\sqrt{n+1}}< 2\)
Ta có đpcm.
Đúng 1
Bình luận (0)