Ban su dung phuong phap: Dua ton duoi can ve binh phuong
1^3+2^3= 1+8= 9 = 3^2= (1+2)^2 Vay √13+23=1+2
1^3+2^3+3^3= 9+27=36=6^2=(1+2+3)^2 Vay √13+23+33=1+2+3
1^3+2^3+3^3+4^3=36+64=100=10^2=(1+2+3+4)^2. Vay √13+23+33+43=1+2+3+4
Ban su dung phuong phap: Dua ton duoi can ve binh phuong
1^3+2^3= 1+8= 9 = 3^2= (1+2)^2 Vay √13+23=1+2
1^3+2^3+3^3= 9+27=36=6^2=(1+2+3)^2 Vay √13+23+33=1+2+3
1^3+2^3+3^3+4^3=36+64=100=10^2=(1+2+3+4)^2. Vay √13+23+33+43=1+2+3+4
. Chứng minh đẳng thức
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}=\sqrt{2}-1\) b) \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}=1\)
Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
Chứng minh :
\(\sqrt{1^3+2^3}\) = 1+2
\(\sqrt{1^3+2^3+3^3}=1+2+3\)
\(\sqrt{1^3+2^3+3^3+4^{3^{ }}}=1+2+3+4\)
Chứng minh các bất đẳng thức sau:
\(\frac{\sqrt{\sqrt{2}+\sqrt{2}}+\sqrt{3}\sqrt{2-\sqrt{2}}}{4}< 0,8\)
(\(\sqrt{\sqrt{3}}+\sqrt{\sqrt{5}}+\sqrt{\sqrt{7}}\))-(\(\sqrt{3}+\sqrt{5}+\sqrt{7}\)) < 3
\(\sqrt{\sqrt{17+12\sqrt{2}}-\sqrt{2}}>\sqrt{3}-1\)
Chứng minh đẳng thức sau:
\(\dfrac{1}{1+\sqrt{2}}\)+\(\dfrac{1}{\sqrt{2}+\sqrt{3}}\)+...+\(\dfrac{1}{\sqrt{99}+\sqrt{100}}\)=9
Bài 1:Chứng minh các đẳng thức:
a) \(\sqrt{5}+\sqrt{3}=\sqrt{8+2\sqrt{3}}\)
b)\(\sqrt{5}+2=\sqrt{9+4\sqrt{5}}\)
1.Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
2.Chứng minh: A= \(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
thực hiện phép tính:
1, \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2,\(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}+\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
4,\(\dfrac{3-\sqrt{3}}{2\sqrt{3}-1}+\dfrac{3+\sqrt{3}}{2\sqrt{3}-1}\)
5,\(\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
Rút gọn biểu thức:
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}\)
\(B=\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)
\(C=\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(1:\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(D=2\sqrt{50}-\frac{1}{\sqrt{2}-1}+4\sqrt{\frac{9}{2}}-\sqrt{3-2\sqrt{2}}\)