a) Rút gọn: \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
b) CM: A = \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\) là 1 số nguyên.
B = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Câu 1 : Rút gọn biểu thức
a, \(\frac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\frac{2}{3}\sqrt{12}.\)b, \(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\frac{3}{3+3\sqrt{6}}.\)
c\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\)Với a>0;b>0
Bài 1 Rút gọn các biểu thức
a, \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\) với b>0
b,\(\frac{3+\sqrt{4}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
c,\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
d, A=\(\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{29-6\sqrt{20}}}}\)
e, B=\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Thu gọn biểu thức
a, A = \(\frac{2\sqrt{3-\sqrt{3+\sqrt{3+\sqrt{48}}}}}{\sqrt{6}-2}\)
b, B = \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-b}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+b}\)
a) Rút gọn Q
Rút gọn
A=\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
B= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
C= \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{16}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{5+\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Mn lm đc câu nào thì đc, hết càng tốt <3
cho biểu thức \(P=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{\sqrt{a}}\right):\left(\frac{2a+\sqrt{a}-1}{1-a}+\frac{2a\sqrt{a}+a-\sqrt{a}}{1+a\sqrt{a}}\right)\)
a. rút gọn P KQ=\(\frac{1-\sqrt{a}+a}{\sqrt{a}}\)
b. tính P khi \(a=\frac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13-\sqrt{48}}}}}+1\) KQ =7/3
c. tìm x để P>x
lm hooj t câu c vs câu a,b, t lm hết r
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. \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}\)
a) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\sqrt{\frac{x}{y}}\)
c) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
d) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(\sqrt{1}-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
e) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
Rút gọn các biểu thức
a, \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
b, \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
c, \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
d, \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
a) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
b) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}}{\sqrt{y}}\) (Bạn tự thêm đk)
d) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\) (Bạn tự thêm đk)