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12 tháng 6 2017

a)   \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)

Vì   \(x< -\frac{1}{2}\)nên   \(\left|2x+1\right|=-\left(2x+1\right)\)

\(\Rightarrow2x+2x+1=4x+1\)

b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)

Khi   \(x\ge\frac{2}{3}\)thì   \(\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)

Khi     \(x< \frac{2}{3}\)  thì  \(\left|3x-2\right|=2-3x\)

\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)

c)  \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)

Đặt   \(\sqrt{a}=x\)  ta được :  \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do  \(a\ge0\))

d)  \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)

TK NKA !!!

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)

\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)

_Minh ngụy_

2 tháng 7 2019

\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)

\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)

\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)

\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)

\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)

2 tháng 7 2019

\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé 

\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)

\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)

24 tháng 7 2017

a, dk \(x\ge0.x\ne1\)

\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)

 =\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)

phan b,c ban tu lam not nhe dai lam mk ko lam dau  mk co vc ban rui

15 tháng 8 2017

a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)

\(=3\sqrt{xy}\)

b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)

15 tháng 8 2017

a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)

29 tháng 7 2018

a)  \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)

b)  \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)

d)  \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)

21 tháng 7 2018

a. =\(\frac{x\sqrt{xy}+y\sqrt{x^2}-x\sqrt{y^2}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}\)
=\(\frac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{\sqrt{xy}\left(x-y\right)}{\sqrt{xy}}\)=\(x-y\)
b. =\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x-1}}\)=\(x+\sqrt{x}+1\)

10 tháng 1 2019

a/ \(P=\frac{1}{\sqrt{xy}}\)

b/ \(x^3=8-6x\)

\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)