Chứng minh:
a) (\(\sqrt{3}-1\))\(^2\) = \(4-2\sqrt{3}\)
b) \(\sqrt{4-2\sqrt{ }3}-\sqrt{3=-1}\)
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.
\(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}\\ =\sqrt{x-4-2\sqrt{x-4}+1}-\sqrt{x-4-4\sqrt{x-4}+4}\\ =\sqrt{\left(\sqrt{x-4}-1\right)^2}-\sqrt{\left(\sqrt{x-4}-2\right)^2}\\ =\sqrt{x-4}-1-\sqrt{x-4}+2\\ =1\left(đpcm\right)\)\
Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
1.Chứng minh:\(\dfrac{a+\sqrt{2+\sqrt{5}.}\sqrt{\sqrt{9-4\sqrt{5}}}}{3\sqrt{2-\sqrt{5}}.\sqrt[3]{\sqrt{9+4\sqrt{5}-}3\sqrt{a^2}+\sqrt[3]{a}}}\)=\(-\sqrt[3]{a}-1\)
2.Rút gọn: \(\left(\dfrac{a^3\sqrt[]{a}-2a^3\sqrt{b}+\sqrt[3]{a^2}-\sqrt[3]{b}}{\sqrt[3]{a^2-\sqrt[3]{ab}}}+\dfrac{\sqrt[3]{a^2b}-\sqrt[3]{ab^2}}{\sqrt[3]{a}-\sqrt[3]{b}}\right)1\dfrac{1}{\sqrt[3]{a^2}}\)
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 đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
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. \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) ; \(\frac{7}{5}< \frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}< \frac{29}{30}\)
chứng minh rằng
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}=\frac{2}{\sqrt[]{x}}\)
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
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)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
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