\(4\sqrt{1-a}=a+6-3\sqrt{1-a^2}+5\sqrt{1+a}\)
Những câu hỏi liên quan
Tính :
a) dfrac{5+2sqrt{5}}{sqrt{5}}+dfrac{3+sqrt{3}}{sqrt{3}}-left(sqrt{5}+sqrt{3}right)
b) left(dfrac{1}{2-sqrt{5}}+dfrac{2}{sqrt{5}+sqrt{3}}right):dfrac{1}{sqrt{21+12sqrt{3}}}
c) dfrac{sqrt{2}+sqrt{3}+sqrt{4}}{sqrt{2}+sqrt{3}+sqrt{6}+sqrt{8}+4}
d) sqrt{21-6sqrt{6}}+sqrt{9+2sqrt{18}}-2sqrt{6+3sqrt{3}}
e) sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
f) dfrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)left(sqrt{5-2sqrt{6}}right)}{9sqrt{3}-11sqrt{2}}
g) left(dfrac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}righ...
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Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
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Rút gọn:
A dfrac{4+sqrt{7}}{3sqrt{2}+sqrt{4+sqrt{7}}}+dfrac{4-sqrt{7}}{3sqrt{2}-sqrt{4-sqrt{7}}}
B dfrac{3sqrt{2}+sqrt{11}}{sqrt{2}+sqrt{6+sqrt{11}}}+dfrac{3sqrt{2}-sqrt{11}}{sqrt{2}-sqrt{6-sqrt{11}}}+18
C dfrac{1}{sqrt{3}+sqrt{5}}+dfrac{1}{sqrt{5}+sqrt{7}}+...+dfrac{1}{sqrt{2n+1}+sqrt{2n+3}}với n thuộc N*
D left(sqrt{3}+1right)left(sqrt{5}-1right)left(sqrt{15}-1right)left(7-2sqrt{3}+sqrt{5}right)
Edfrac{left(4+sqrt{3}right)}{sqrt[]{1}+sqrt{3}}+dfrac{left(8+sqrt{15}right)}{sqrt{3}+sqrt...
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Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1
Bài 1: Rút gọn biểu thức:Afrac{a^3-3a+left(a^2-1right)sqrt{a^2-4}-2}{a^3-3a+left(a^2-1right)sqrt{a^2-4}+2}left(a2right)Bsqrt{frac{1}{a^2+b^2}+frac{1}{left(a+bright)^2}+sqrt{frac{1}{a^4}+frac{1}{b^4}+frac{1}{left(a^2+b^2right)^2}}}left(abne0right)Bài 2: Tính giá trị của biểu thức:Efrac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{2sqrt{3}+3sqrt{2}}+frac{1}{3sqrt{4}+4sqrt{3}}+...+frac{1}{2017sqrt{2018}+2018sqrt{2017}}Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyênAleft(sqrt{57}+3sqrt{6}+sqrt{38}...
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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
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
1/ Tính:
a) frac{sqrt{6+sqrt{11}}-sqrt{7-sqrt{33}}}{sqrt{6}+sqrt{2}}
b) frac{5sqrt{3}-3sqrt{5}}{sqrt{5}-sqrt{3}}+frac{2}{4+sqrt{15}}-frac{5sqrt{5}+3sqrt{3}}{sqrt{5}+sqrt{3}}
2/ Rút Gọn: với a ≥ 0, a ≠ 1
Bleft(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)left(frac{1+sqrt{a}}{a-1}right)^2
3/ Cho biểu thức: A frac{sqrt{x}+2}{sqrt{x}-3}-frac{sqrt{x}+1}{sqrt{x}-2}+frac{3-3sqrt{x}}{x-5sqrt{x}+6}
a) Tìm điều kiện xác định của A
b) Rút gọn A
c) Tìm x để A -1
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1/ Tính:
a) \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)
b) \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
2/ Rút Gọn: với a ≥ 0, a ≠ 1
B=\(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)
3/ Cho biểu thức: A = \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)
a) Tìm điều kiện xác định của A
b) Rút gọn A
c) Tìm x để A < -1
Bài 1: Tính
a) Ta có: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{\sqrt{12}+2}\)
\(=\frac{\sqrt{11+2\cdot\sqrt{11}\cdot1+1}-\sqrt{11-2\cdot\sqrt{11}\cdot\sqrt{3}+3}}{2\sqrt{3}+2}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}}{2\sqrt{3}+2}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-\sqrt{3}\right|}{2\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{11}+1-\left(\sqrt{11}-\sqrt{3}\right)}{2\left(1+\sqrt{3}\right)}\)(Vì \(\left\{{}\begin{matrix}\sqrt{11}>1>0\\\sqrt{11}>\sqrt{3}\end{matrix}\right.\))
\(=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt{3}}{2\left(1+\sqrt{3}\right)}\)
\(=\frac{1+\sqrt{3}}{2\left(1+\sqrt{3}\right)}=\frac{1}{2}\)
b) Ta có: \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\frac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(8-\sqrt{15}\right)}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-\left(8-\sqrt{15}\right)\)
\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-8+\sqrt{15}\)
\(=2\sqrt{15}-8+\frac{2}{4+\sqrt{15}}\)
\(=\frac{2\sqrt{15}\left(4+\sqrt{15}\right)}{4+\sqrt{15}}-\frac{8\left(4+\sqrt{15}\right)}{4+\sqrt{15}}+\frac{2}{4+\sqrt{15}}\)
\(=\frac{8\sqrt{15}+30-32-8\sqrt{15}+2}{4+\sqrt{15}}\)
\(=\frac{0}{4+\sqrt{15}}=0\)
Bài 2: Rút gọn
Ta có: \(B=\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)
\(=\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\cdot\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)^2\)
\(=\left(1-\sqrt{a}+a-\sqrt{a}\right)\cdot\left(\frac{1}{\sqrt{a}-1}\right)^2\)
\(=\left(a-2\sqrt{a}+1\right)\cdot\frac{1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2}=1\)
Bài 3:
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b) Ta có: \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{3-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)+3-3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-1-x+2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{1}{3-\sqrt{x}}\)
c) Để A<-1 thì A+1<0
\(\Leftrightarrow\frac{1}{3-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\frac{-1}{\sqrt{x}-3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>4\\\sqrt{x}< 3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 4\\\sqrt{x}>3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\Leftrightarrow9< x< 16\)
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1. CHỨNG MINH ĐẲNG THỨCa. text{[}3+2sqrt{6}-sqrt{33}text{]}cdottext{[}sqrt{22}+sqrt{6}+4text{]}24b. text{[}frac{1}{5-2sqrt{6}}+frac{2}{5+2sqrt{6}}text{]}cdottext{[}15+2sqrt{6}text{]}c.text{[}frac{4}{3}cdotsqrt{3}+sqrt{2}+sqrt{3frac{1}{3}}text{]}cdottext{[}sqrt{1,2}+sqrt{2}-4sqrt{frac{1}{5}}text{]}4d. sqrt{text{[}1-sqrt{1989}text{]}^2}cdotsqrt{1990+2sqrt{1989}}1988e. frac{a-sqrt{ab}+b}{asqrt{a}+bsqrt{b}}-frac{1}{a-b}frac{sqrt{a}-sqrt{b}-1}{a-b}với a0;b0và ane b
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1. CHỨNG MINH ĐẲNG THỨC
a. \(\text{[}3+2\sqrt{6}-\sqrt{33}\text{]}\cdot\text{[}\sqrt{22}+\sqrt{6}+4\text{]}=24\)
b. \(\text{[}\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\text{]}\cdot\text{[}15+2\sqrt{6}\text{]}\)
c.\(\text{[}\frac{4}{3}\cdot\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\text{]}\cdot\text{[}\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\text{]}=4\)
d. \(\sqrt{\text{[}1-\sqrt{1989}\text{]}^2}\cdot\sqrt{1990+2\sqrt{1989}}=1988\)
e. \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)với \(a>0;b>0\)và \(a\ne b\)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
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Bài 1 : Tính :
a) dfrac{1}{5+2sqrt{6}}-dfrac{1}{5-2sqrt{6}}
b) sqrt{6+2sqrt{5}}-dfrac{sqrt{15}-sqrt{3}}{sqrt{3}}
c) dfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}:dfrac{1}{sqrt{16}}
d) dfrac{3+2sqrt{3}}{sqrt{3}}+dfrac{2+sqrt{2}}{1+sqrt{2}}-dfrac{1}{2-sqrt{3}}
e) dfrac{4}{1+sqrt{3}}-dfrac{sqrt{15}+sqrt{3}}{1+sqrt{5}}
f) left(dfrac{1}{2-sqrt{5}}+dfrac{2}{sqrt{5}-sqrt{3}}right):dfrac{1}{sqrt{21-12sqrt{3}}}
Bài 2 : Rút gọn :
a) dfrac{a+b-2sqrt{ab}}{sqrt{a}-sqrt{b}}:dfrac{1}{sqrt{a}+sqrt{b}}
b) l...
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Bài 1 : Tính :
a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
f) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}-\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
Bài 2 : Rút gọn :
a) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
c) \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
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bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
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Cho a, b là những số thực dương. Rút gọn các biểu thức sau:
a) dfrac{a^{dfrac{4}{3}}(a^{dfrac{-1}{3}}+a^{dfrac{2}{3}})}{a^{dfrac{1}{4}}(a^{dfrac{3}{4}}+a^{dfrac{-1}{4}})}
b) dfrac{b^{dfrac{1}{5}} (sqrt[5]{b^4}-sqrt[5]{b^{-1}})}{b^{dfrac{2}{3}}(sqrt[3]{b}-sqrt[3]{b^{-2}})}
c) dfrac{a^{dfrac{1}{3}}b^{dfrac{-1}{3}}-a^{dfrac{-1}{3}}b^{dfrac{1}{3}}}
{sqrt[3]{a^2}-sqrt[3]{b^2}}
d) dfrac{a^{dfrac{1}{3}} sqrt{b}+b^{dfrac{1}{3}} sqrt{a}}
{sqrt[6]{a}+sqrt[6]{b}}
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Cho a, b là những số thực dương. Rút gọn các biểu thức sau:
\(a)\ \dfrac{a^{\dfrac{4}{3}}(a^{\dfrac{-1}{3}}+a^{\dfrac{2}{3}})}{a^{\dfrac{1}{4}}(a^{\dfrac{3}{4}}+a^{\dfrac{-1}{4}})}\)
\(b)\ \dfrac{b^{\dfrac{1}{5}} (\sqrt[5]{b^4}-\sqrt[5]{b^{-1}})}{b^{\dfrac{2}{3}}(\sqrt[3]{b}-\sqrt[3]{b^{-2}})}\)
\(c)\ \dfrac{a^{\dfrac{1}{3}}b^{\dfrac{-1}{3}}-a^{\dfrac{-1}{3}}b^{\dfrac{1}{3}}}
{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)
\(d)\ \dfrac{a^{\dfrac{1}{3}} \sqrt{b}+b^{\dfrac{1}{3}} \sqrt{a}}
{\sqrt[6]{a}+\sqrt[6]{b}}\)
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
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Rút gọn: ( 2,5 Điểm )A dfrac{sqrt{6+2sqrt{5}}}{sqrt{5}+1}+ dfrac{sqrt{5-2sqrt{6}}}{sqrt{3}-sqrt{2}}B dfrac{3}{sqrt{5}-2}+ dfrac{4}{sqrt{6}+sqrt{2}}+ dfrac{1}{sqrt{6}+sqrt{5}}C dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{99}+sqrt{100}}D dfrac{1}{2-sqrt{3}}+sqrt{7-4sqrt{3}}E sqrt{dfrac{3sqrt{3}-4}{2sqrt{3}+1}}-sqrt{dfrac{sqrt{3}+4}{5-2sqrt{3}}}F dfrac{1}{2+sqrt{3}}+dfrac{sqrt{2}}{sqrt{6}}-dfrac{2}{3+sqrt{3}}
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Rút gọn: ( 2,5 Điểm )
A= \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)+ \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
B= \(\dfrac{3}{\sqrt{5}-2}\)+ \(\dfrac{4}{\sqrt{6}+\sqrt{2}}\)+ \(\dfrac{1}{\sqrt{6}+\sqrt{5}}\)
C = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
D= \(\dfrac{1}{2-\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
E = \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
F = \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
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Rút gọn
a, frac{2sqrt{3-1}}{sqrt{15}}-frac{2-sqrt{5}}{sqrt{3}}-frac{4sqrt{15}-10sqrt{3}}{15}
b, left(sqrt{a}-frac{1}{sqrt{a}}right)left(frac{sqrt{a}+1}{sqrt{a-1}}+frac{sqrt{a-1}}{sqrt{a}+1}right)
c, sqrt{4+sqrt{7}-sqrt{4-sqrt{7}}}
d, 6+2sqrt{2}.3-sqrt{4+sqrt{2sqrt{3}}}
e, sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}
Help me !!!
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Rút gọn
a, \(\frac{2\sqrt{3-1}}{\sqrt{15}}-\frac{2-\sqrt{5}}{\sqrt{3}}-\frac{4\sqrt{15}-10\sqrt{3}}{15}\)
b, \(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a-1}}+\frac{\sqrt{a-1}}{\sqrt{a}+1}\right)\)
c, \(\sqrt{4+\sqrt{7}-\sqrt{4-\sqrt{7}}}\)
d, \(6+2\sqrt{2}.3-\sqrt{4+\sqrt{2\sqrt{3}}}\)
e, \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
Help me !!!