Chứng minh \(\sqrt{1^3+2^3+3^3+4^3}=1+2+3+4\)
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1) Chứng minh đẳng thức \(\left(\frac{3\sqrt{2}-\sqrt{6}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}=-\frac{4}{3}\)
2) Chứng minh \(\sqrt{\sqrt{3}-\sqrt{3-\sqrt{13-4\sqrt{3}}}}=1\)
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
\(\dfrac{3}{2}\)\(\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
rút gọn
D=\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\)\(-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
a)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
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b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)
\(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
D=2
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Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
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Chứng minh:
a, \(\sqrt{\sqrt{3}+2\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt{3}-2\sqrt{\sqrt{3}-1}}=2\)
b, \(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}=1\)
a,
\(\sqrt{\sqrt{3}+2\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt{3}-2\sqrt{\sqrt{3}-1}}\\ =\sqrt{\sqrt{3}-1+2\sqrt{\sqrt{3}-1}+1}+\sqrt{\sqrt{3}-1-2\sqrt{\sqrt{3}-1}+1}\\ =\sqrt{\left(\sqrt{\sqrt{3}-1}+1\right)^2}+\sqrt{\left(1-\sqrt{\sqrt{3}-1}\right)^2}\\ =\sqrt{\sqrt{3}-1}+1+1-\sqrt{\sqrt{3}-1}\\ =2\)
b.
\(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}\\ =\sqrt{x-4-2\sqrt{x-4}+1}-\sqrt{x-4-4\sqrt{x-4}+4}\\ =\sqrt{\left(\sqrt{x-4}-1\right)^2}-\sqrt{\left(\sqrt{x-4}-2\right)^2}\\ =\sqrt{x-4}-1-\sqrt{x-4}+2\\ =1\left(đpcm\right)\)\
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HD
30 tháng 10 2023 lúc 19:48
Cho \(x=\dfrac{\sqrt{2}-1}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{225}-\sqrt{224}}{224+225}\) . Chứng minh rằng \(x< \dfrac{7}{15}\) .
Chứng minh :
\(\sqrt{1^3+2^3}\) = 1+2
\(\sqrt{1^3+2^3+3^3}=1+2+3\)
\(\sqrt{1^3+2^3+3^3+4^{3^{ }}}=1+2+3+4\)
\(\sqrt{1^3+2^3}=1+2\)
\(\Leftrightarrow\sqrt{1+8}=3\)
\(\Leftrightarrow\sqrt{9}=3\)
mà \(\sqrt{9}=\sqrt{3^2}=\left|3\right|=3\)
\(\Leftrightarrow3=3\)
\(\Rightarrow\sqrt{1^3+2^3}=1+2\)
mấy bài khác chị giải tương tự là ra.
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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)
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Chứng minh đẳng thức (sqrt(4 - 2sqrt(3)))/(1 + sqrt(2)) / ((sqrt(2) - 1)/(sqrt(3) + 1)) = 2
\(=\dfrac{\sqrt{3}-1}{\sqrt{2}+1}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2}{1}=2\)
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