a) CMR \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0\)và \(a\ne1\).
b) CMR \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}=1\)
Những câu hỏi liên quan
Chứng minh các đẳng thức saua) left(frac{2sqrt{6}-sqrt{3}}{2sqrt{2}-1}+frac{5+2sqrt{5}}{2+sqrt{5}}right)left(sqrt{5}-sqrt{3}right)b) frac{a-b}{b^2}sqrt{frac{a^2b^4}{a^2-2ab+b^2}}-a(Với ba0c)left(sqrt{a}+frac{1-asqrt{a}}{1-sqrt{a}}right)left(frac{1-sqrt{a}}{1-a}right)^21với age0,a khác 1d) left(frac{3sqrt{5}-sqrt{15}}{sqrt{27}-3}+frac{2sqrt{5}}{sqrt{3}}right)40sqrt{15}600e) left(1+frac{x+sqrt{x}}{sqrt{x}+1}right)left(1-frac{x-sqrt{x}}{sqrt{x}-1}right)1-xvới xge0;xne1
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Chứng minh các đẳng thức sau
a) \(\left(\frac{2\sqrt{6}-\sqrt{3}}{2\sqrt{2}-1}+\frac{5+2\sqrt{5}}{2+\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
b) \(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=-a\)(Với b<a<0
c)\(\left(\sqrt{a}+\frac{1-a\sqrt{a}}{1-\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)với a\(\ge0\),a khác 1
d) \(\left(\frac{3\sqrt{5}-\sqrt{15}}{\sqrt{27}-3}+\frac{2\sqrt{5}}{\sqrt{3}}\right)40\sqrt{15}=600\)
e) \(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)với x\(\ge0;x\ne1\)
Rút gọn:
a) \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\left(x\ge0,x\ne1\right)\)
b) \(B=\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\left(a\ge0,a\ne2,a\ne4\right)\)
c) \(C=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
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C/m biểu thứca)left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}+sqrt{b}}-sqrt{ab}right)left(frac{sqrt{a}+sqrt{b}}{a-b}right)1(a,b0,ane0b)frac{a-b-2sqrt{ab}}{sqrt{a}-sqrt{b}}:frac{1}{sqrt{a}+sqrt{b}}a-bleft(a,b0,ane bright)c)left(2+frac{a-sqrt{a}}{sqrt{a}-1}right)left(2-frac{a+sqrt{a}}{sqrt{a}+1}right)4-aleft(a0,ane1right)d)left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)left(1-aright)^2left(age0,ane1right)Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
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C/m biểu thức
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)=1\)(a,b>0,a\(\ne\)0
b)\(\frac{a-b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\left(a,b>0,a\ne b\right)\)
c)\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\left(a>0,a\ne1\right)\)
d)\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1-a\right)^2\left(a\ge0,a\ne1\right)\)
Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
Bài 1:Rút gọna,frac{2}{sqrt{3}-1}-frac{2}{sqrt{3}+1}b,frac{2}{5-sqrt{3}}+frac{3}{sqrt{6}+sqrt{3}}c,left(1+frac{a+sqrt{a}}{1+sqrt{a}}right)timesleft(1-frac{a-sqrt{a}}{sqrt{a}-1}right)+aleft(ane1;age0right)Bài 2: Rút gọn biểu thứcPfrac{2sqrt{x}-9}{left(sqrt{x}-2right)left(sqrt{x}-3right)}-frac{sqrt{x}+3}{sqrt{x}-2}-frac{2sqrt{x}+1}{3-sqrt{x}}
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Bài 1:Rút gọn
\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(b,\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(c,\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)\times\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\left(a\ne1;a\ge0\right)\)
Bài 2: Rút gọn biểu thức
\(P=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
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Cho B=\(\left(\frac{a\sqrt{a}-3}{a-2\sqrt{a}-3}-\frac{2\left(\sqrt{a}-3\right)}{\sqrt{a}+1}+\frac{\sqrt{a}+3}{3-\sqrt{a}}\right):\left(\frac{a+8}{a-1}\right)\)
Rút gọn A với a\(\ge0;a\ne9;a\ne1\)
CHỨNG MINH
a) frac{left(sqrt{a}+1right)^2-4sqrt{a}}{sqrt{a}-1}+frac{a+sqrt{a}}{sqrt{a}}2sqrt{a} left(a0;ane1right)
b) frac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-left(sqrt{x}-sqrt{y}right)^2sqrt{xy} left(xge0;yge0right)
c) frac{asqrt{b}-bsqrt{a}}{sqrt{ab}}:frac{a-b}{sqrt{a}-sqrt{b}}1 left(a0;b0;ane bright)
d) left[frac{left(sqrt{a}-sqrt{b}right)^2+4sqrt{ab}}{sqrt{a}+sqrt{b}}-frac{asqrt{b}-bsqrt{a}}{sqrt{ab}}right]:sqrt{b}2 left(a0;b0right)
Giúp mình với, cảm ơn mn 3
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CHỨNG MINH
a) \(\frac{\left(\sqrt{a}+1\right)^2-4\sqrt{a}}{\sqrt{a}-1}+\frac{a+\sqrt{a}}{\sqrt{a}}=2\sqrt{a}\) \(\left(a>0;a\ne1\right)\)
b) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\) \(\left(x\ge0;y\ge0\right)\)
c) \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\frac{a-b}{\sqrt{a}-\sqrt{b}}=1\) \(\left(a>0;b>0;a\ne b\right)\)
d) \(\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\right]:\sqrt{b}=2\) \(\left(a>0;b>0\right)\)
Giúp mình với, cảm ơn mn <3
cau c í mk thấy bn chép sai đề nên mk sửa lại đề rồi bạn xem lại đề rồi so với bài làm của mk nha có j ko hiểu thì ib mk nha
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\(a)VT = \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\ = \dfrac{{a + 2\sqrt a + 1 - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a }}\\ = \dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)}} + \sqrt a + 1\\ = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a - 1}} + \sqrt a + 1\\ = \sqrt a - 1 + \sqrt a + 1\\ = 2\sqrt a = VP (đpcm) \)
\(b)VT = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt {xy} + y} \right)\\ = x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\ = \sqrt {xy} (đpcm)\\ c)VT = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\ = \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \sqrt a - \sqrt b .\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\\ = \dfrac{{a - b}}{{a - b}} = 1 (đpcm)\\ d)VT = \left[ {\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}} \right]:\sqrt b \\ = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}:\sqrt b \\ = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \left( {\sqrt a - \sqrt b } \right):\sqrt b \\ = \sqrt a + \sqrt b - \sqrt a + \sqrt b :\sqrt b \\ = \dfrac{{2\sqrt b }}{{\sqrt b }} = 2 (đpcm) \)
Câu c đề sai (đã sửa)
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Rút gọn:
a) \(B=\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\left(a\ge0,a\ne2,a\ne4\right)\)
b) \(C=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
Rút gọn biểu thức:
a) \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\frac{2}{a+1}\right)^2\) với \(a>0,a\ne1\)
a) Ta có: \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\left|\sqrt{5}-1\right|}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)(Vì \(\sqrt{5}>1\))
\(=\left(2\sqrt{4+\sqrt{5}-1}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{5+2\cdot\sqrt{5}\cdot1+1}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left|\sqrt{5}+1\right|\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)\)
\(=2\cdot\left(5-1\right)\)
\(=2\cdot4=8\)
b) Ta có: \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\frac{2}{a+1}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2+\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\right)\cdot\left(\frac{a+1-2}{a+1}\right)^2\)
\(=\frac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2a+2}{\left(a-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2\left(a+1\right)\cdot\left(a-1\right)}{\left(a+1\right)^2}\)
\(=\frac{2a-2}{a+1}\)
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A=\(\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right):\frac{\sqrt{a}-2}{a+1}\left(a\ge0;a\ne1\right)\)
Rút gọn
A= (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\)) : \(\frac{\sqrt{a}-2}{a+1}\)
<=> (\(\frac{1}{\sqrt{a-1}}\) - \(\frac{2\sqrt{a}}{\left(a\sqrt{a-1}\right)-\sqrt{a}\left(\sqrt{a}-1\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1-\sqrt{a}\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{a+1}{\left(\sqrt{a}-1\right)\left(a+1\right)}\).\(\frac{a+1}{\sqrt{a}+2}\)
<=> \(\frac{a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
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