Trực căn thức ở mẫu
a)\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}}\) b)\(\frac{2\sqrt{5}}{\sqrt{3}}\) c)\(\frac{1}{2\sqrt{32}}\) d)\(\frac{x+\sqrt{y}}{x\sqrt{y}}\) e)\(\frac{3\sqrt{2}-1}{3-\sqrt{8}}\) f)\(\frac{3-3\sqrt{2}}{2\sqrt{2}-3\sqrt{6}}\)
g)\(\frac{xy}{\sqrt{x}-\sqrt{y}}\)
1. Tính:
a) A= \(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)(dấu căn đầu tiên là của cả biểu thức)
b) B= \(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
2. Cho:
A= \(\left(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right):\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)với x>0, y>0
a) Rút gọn A
b) Cho xy=16. Tìm x,y để A có GTNN. Tìm Gt đó
\(A=\left(\sqrt{5}-\sqrt{2}\right)^2-\frac{9}{\sqrt{10}-1}+\sqrt{90}\)\(B=\sqrt{2}\left(3\sqrt{2}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\)\(C=\left(\frac{5-\sqrt{5}}{\sqrt{5}-1}-\frac{\sqrt{5}+1}{5+\sqrt{5}}\right):\frac{\sqrt{5}+1}{\sqrt{5}}\)\(D=\frac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}:\frac{x+2\sqrt{xy}+y}{\left(\sqrt{x}+\sqrt{y}\right)^3\left(x+y\right)}vớix,y>0\)
TÍNH HOẶC RÚT GỌN
1. Trục căn thức ở mẫu :
a) \(\frac{1+\sqrt{2}}{1-\sqrt{2}}\) b) \(\frac{1}{3\sqrt{2}-2\sqrt{3}}\) c) \(\frac{1-x^2}{1-\sqrt{x}}\) d) \(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\) e) \(\frac{3\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)
f) \(\frac{4}{1+\sqrt{2}+\sqrt{3}}\) g) \(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\)
bài 1: rút gọn:
C=\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
bài 2 :rút gọn
E=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
1. RÚT GỌN BIỂU THỨC
a. \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
b. \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
c. \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
d. \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
e. \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}\)
Rút gọn biểu thức sau
a/ A=\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\frac{x-y}{\sqrt{x}-\sqrt{y}}\)Với x>0 ; y>0 ;x#y
b/ B=\(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
c/ C=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
d/ D=\(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}\)
đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)
ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)
\(\Leftrightarrow x^5-x^2\ge3x-3\)
cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)
\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)
\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)
áp dụng bunhia ta có:
\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)
cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)
đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)
\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)
\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)
\(\Rightarrow P\le1\)
1) Khử mẫu các biểu thức dưới dấu căn rồi thực hiện phép tính:
\(2\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-\sqrt{\frac{1}{15}}\)
2) Trục căn thức ở mẫu:
a) \(\frac{9}{\sqrt{3}}\)
b) \(\frac{12}{3-\sqrt{3}}\)
c) \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
d) \(\frac{7\sqrt{3}-5\sqrt{11}}{8\sqrt{3}-7\sqrt{11}}\)
e) \(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
f) \(\frac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}\)
g) \(\frac{1}{1+\sqrt{2}-\sqrt{3}}\)
h) \(\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\)