a) x + y - \(2\sqrt{xy}\) - 9
b) 1 + \(\sqrt{13}+\sqrt{5}+\sqrt{65}\)
Rút gọn :
a) \(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\)
b) \(\sqrt{\left(\sqrt{5}-1\right).\sqrt{13-\sqrt{69-28\sqrt{5}}}}\)
c) \(\dfrac{\sqrt{3+\sqrt{5}}.\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}\)
\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)
\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)
\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)
Câu a
\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
\(A=\left(\sqrt{5}-\sqrt{2}\right)^2-\frac{9}{\sqrt{10}-1}+\sqrt{90}\)\(B=\sqrt{2}\left(3\sqrt{2}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\)\(C=\left(\frac{5-\sqrt{5}}{\sqrt{5}-1}-\frac{\sqrt{5}+1}{5+\sqrt{5}}\right):\frac{\sqrt{5}+1}{\sqrt{5}}\)\(D=\frac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}:\frac{x+2\sqrt{xy}+y}{\left(\sqrt{x}+\sqrt{y}\right)^3\left(x+y\right)}vớix,y>0\)
TÍNH HOẶC RÚT GỌN
1, \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
2, \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)
3, \(\dfrac{9\sqrt{a}-b\sqrt{5}}{\sqrt{a}-\sqrt{5}}+\sqrt{ab}\)
4, \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)\)
5, \(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
1. \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\sqrt{a}-2\)
= 0
2: \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)
4: \(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\dfrac{1}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}=\sqrt{a}+1\)
Bài 1 :Tinh giá trị của các biểu thức sau :( lấy 05 chữ số thập phân sau dấu phẩy)
a) Tìm x số thực thỏa mãn:
\(\frac{x}{2+\frac{x}{2+\frac{x}{2+\frac{x}{\sqrt{1+x}+1}}}}=2012\)
b)\(P=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
c)\(C=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}\times\frac{1-x}{\left(1-\sqrt{x}\right)\sqrt{y}}\)khi\(x=2,47839;y=\sqrt{7-4\sqrt{3}}\)
Bài 2. Cho A=\(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\) :\([\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{1}{xy+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)]\)
Bạn cần làm gì với biểu thức này?
1.Rút gọn:
a) \(A=\sqrt{2+\sqrt{3}.}\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
b) \(B=\left(\frac{\sqrt{x}}{\sqrt{xy}-y}-\frac{\sqrt{y}}{\sqrt{xy}-x}\right).\left(x\sqrt{y}-y\sqrt{x}\right)\)
c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)
a) \(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(A=\sqrt{\left(2+\sqrt{3}\right)\left(\sqrt{2+\sqrt{3}}+2\right)\left(-\sqrt{2+\sqrt{3}}+2\right)}\)
\(A=\sqrt{1}\)
\(A=1\)
b)\(B=\left(\frac{\sqrt{x}}{\sqrt{xy}-y}-\frac{\sqrt{y}}{\sqrt{xy}-x}\right).\left(x\sqrt{y}-y\sqrt{x}\right)\)
\(B=\frac{\sqrt{xy}}{\sqrt{xy}-y}x\sqrt{y}+\frac{\sqrt{x}}{\sqrt{xy}-y}y\sqrt{x}+\left(-\frac{\sqrt{y}}{\sqrt{xy}-x}\right)^2x\sqrt{y}+y\sqrt{x}\)
\(B=x\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{y}+y\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{x}+x\frac{\sqrt{x}}{\sqrt{xy}-x}\sqrt{y}-y\sqrt{x}\frac{\sqrt{y}}{\sqrt{xy}-y}\)
\(B=\frac{-x^{\frac{5}{2}}\sqrt{y}+\sqrt{x}.y^{\frac{5}{2}}}{\left(\sqrt{xy}-y\right)\left(\sqrt{xy}-x\right)}\)
\(B=\frac{\left(\sqrt{x}.y^{\frac{5}{2}}-x^{\frac{5}{2}}\sqrt{y}\right)\left(y+\sqrt{xy}\right)\left(x+\sqrt{xy}\right)}{\left(-y^2+xy\right)\left(-x^2+xy\right)}\)
c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)
\(C=14-6\sqrt{5}+\sqrt{6}-2\sqrt{5}\)
\(C=14-8\sqrt{5}+\sqrt{6}\)
\(C=\sqrt{14-8\sqrt{5}+\sqrt{6}}\)
Rút gọn biểu thức
a) \(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
b)\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
c)\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)
d) \(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)
e)\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y},x,y>0\)
f)\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
g)\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}v\text{ới}a>0,a\ne3\)
\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)
\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)
= - 1
\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}+1}{2}\)
\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)
\(=\dfrac{2\sqrt{2}+2}{\sqrt{3+2\sqrt{2}}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
= 2
\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)
\(=4+9+16+49\)
= 78
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)
\(=\sqrt{x}-\sqrt{y}\)
\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)
\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)
\(=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1\)
= 3
Ta có:
\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)
\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)
\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)
\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)
\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)
Rút gọn
1) \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{ab}}.\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\) với a,b lớn hơn hoặc bằng 0,a khác b.
2) \(\left(2-\frac{7+3\sqrt{7}}{\sqrt{7}+3}\right).\left(2-\frac{5\sqrt{7}-\sqrt{14}}{\sqrt{2}-5}\right)\)
3) \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1-\sqrt{xy}}\right):\left(\frac{x+xy}{1-xy}\right)\)với x,y lớn hơn 0,x,y khác 1
3)\(...=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\right].\frac{1-xy}{x+xy}\)
= \(\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{x\left(1+y\right)}\)= \(\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(1+y\right)}=\frac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}=\frac{2}{\sqrt{x}}\)
\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\)
a) Rút gọn
b) Tính C với x=2-\(\sqrt{3}\); y=2+\(\sqrt{3}\)
Rút gọn:
a) \(\sqrt{27\left(\sqrt{3}-\sqrt{5}\right)^2}\)
b) \(\dfrac{x+\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}\)
a) Ta có: \(\sqrt{27\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=3\sqrt{15}-9\)
c) Ta có: \(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}=x-\sqrt{x}+1\)