CM: \(\sqrt{a+b}+\sqrt{a-b}< 2\sqrt{a},vớia,b,c>0\)
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rút gọn các biểu thức:
a,6sqrt{a}+dfrac{2}{3}sqrt{dfrac{a}{4}}-asqrt{dfrac{9}{a}}+sqrt{7}vớia0
b,5asqrt{25ab^3}sqrt{3}sqrt{12a^3b^3}+9absqrt{9ab}-5bsqrt{81a^3b}vớia,b0
c,sqrt{dfrac{a}{b}}+sqrt{ab}-dfrac{a}{b}sqrt{dfrac{b}{a}}vớia,b0
d,11sqrt{5a}-sqrt{125a}+sqrt{20a}-4sqrt{45a}+9sqrt{a}vớia0
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rút gọn các biểu thức:
a,\(6\sqrt{a}+\dfrac{2}{3}\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{9}{a}}+\sqrt{7}vớia>0\)
b,\(5a\sqrt{25ab^3}\sqrt{3}\sqrt{12a^3b^3}+9ab\sqrt{9ab}-5b\sqrt{81a^3b}vớia,b>0\)
c,\(\sqrt{\dfrac{a}{b}}+\sqrt{ab}-\dfrac{a}{b}\sqrt{\dfrac{b}{a}}vớia,b>0\)
d,\(11\sqrt{5a}-\sqrt{125a}+\sqrt{20a}-4\sqrt{45a}+9\sqrt{a}vớia>0\)
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
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a)\(\sqrt{4\left(a-3\right)^2}vớia\ge3\)
b)\(\sqrt{a^2\left(a+1\right)^2}vớia>0\)
c)\(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}vớia< 0,b\ne0\)
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
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a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
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a,\(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2.b}-\sqrt{a.b^2}\left(Vớia>0,b>0\right)\)
b,\(x-y+\sqrt{x.y^2}-\sqrt{y^3}\left(Vớix>0,y>0\right)\)
a, \(\sqrt{a^3}\)-\(\sqrt{b^3}\)+\(\sqrt{a^2b}\)-\(\sqrt{ab^2}\)
=(\(\sqrt{a^3}\)-\(\sqrt{b^3}\))+(\(\sqrt{a^2b}\)-\(\sqrt{ab^2}\)). =(\(\sqrt{a}\)-\(\sqrt{b}\))(a+\(\sqrt{ab}\)+b)+\(\sqrt{ab}\)(\(\sqrt{a}\)-\(\sqrt{b}\)). =(\(\sqrt{a}\)-\(\sqrt{b}\))(a+\(\sqrt{ab}\)+b+\(\sqrt{ab}\)). =(\(\sqrt{a}\)-\(\sqrt{b}\))(a+2\(\sqrt{ab}\)+b). =(\(\sqrt{a}\)-\(\sqrt{b}\))(\(\sqrt{a}\)+\(\sqrt{b}\))\(^2\) =(a-b)(\(\sqrt{a}\)+\(\sqrt{b}\))
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b, x-y+\(\sqrt{xy^2}\)-y\(^3\) =(x-y)+(\(\sqrt{xy^2}\)-\(\sqrt[3]{y^3}\)). =(\(\sqrt{x}\)-\(\sqrt{y}\))(\(\sqrt{x}\)+\(\sqrt{y}\))+\(\sqrt{y^2}\)(\(\sqrt{ }x\)-\(\sqrt{y}\)). =(\(\sqrt{x}\)-\(\sqrt{y}\))(\(\sqrt{x}\)+\(\sqrt{y}\)+\(\sqrt{y^2}\)). =(\(\sqrt{x}\)-\(\sqrt{y}\))(\(\sqrt{x}\)+\(\sqrt{y}\)+y) (vì y>0).
Rút gon các biểu thức:
a)\(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
b)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
c)\(\sqrt{9\left(3-a\right)^2}vớia>3\)
d)\(\sqrt{a^2.\left(a-2\right)^2}vớia< 0\)
\(\left(a\right)\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
\(\left(b\right) \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2\right)}{\sqrt{2}+\sqrt{3}+2}\\=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\\ =1+\sqrt{2}\)
\(\left(c\right)\sqrt{9\left(3-a\right)^2}vớia>3\\ =\sqrt{9}.\sqrt{\left(3-a\right)^2}\\ =3.\left|3-a\right|\\ =-3\left(3-a\right)vì.a>3\\ =3a-9\)
\(\left(d\right)\sqrt{a^2.\left(a-2\right)^2}vớia< 0\\ =\sqrt{\left[a\left(a-2\right)\right]^2}\\ =\left|a\left(a-2\right)\right|=-a.\left[-\left(a-2\right)\right]=a\left(a-2\right)=a^2-2a\)
Chúc bạn học tốt ! 
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Rút gọn biểu thức:sqrt{frac{2a}{3}}.sqrt{frac{3a}{8}}vớiage0sqrt{5a}.sqrt{45a}-3avớiage04sqrt{16a^6}-6a^3rightarrow kq2THleft(3-aright)^2-sqrt{0,2}.sqrt{180a^4}sqrt{frac{27.left(a-3right)^2}{48}}vớia 3frac{sqrt{63y^3}}{sqrt{7y}}vớiy0frac{sqrt{16a^4b^6}}{sqrt{128a^6b^2}}vớia 0,bne0frac{a-b}{sqrt{a}-sqrt{b}}-frac{sqrt{a^3}+sqrt{b^3}}{a-b}left(age0;bge0;ane bright)frac{2a+sqrt{ab}-3b}{2a-5sqrt{ab}+3b}left(a,bge0;4ane9bright)
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Rút gọn biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)
Rút gọn biểu thức
a) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0) \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\) (a,b ≥ 0; a ≠ b)
b) \(\left(\sqrt{ab}-\sqrt{\frac{a}{b}}+\frac{1}{a}\sqrt{4ab}+\frac{1}{b}\sqrt{\frac{b}{a}}\right):\left(1+\frac{2}{a}-\frac{1}{b}+\frac{1}{ab}\right)vớia,b>0\)
Rút gọn :
a, A = \(\sqrt{27.48\left(1-a^2\right)}vớia>1\)
b, B = \(\frac{1}{a-b}\sqrt{a^4\left(a-b^2\right)}\) với a > b
c, C= \(\sqrt{5a}.\sqrt{45a}-3a\) với a >= 0
tính
a/ \(\sqrt{a^2}vớia=6,5;-0,1;\)
b/ \(\sqrt{a^4}vớia=3;-0,1;\)
c/\(\sqrt{a^6}vớia=-2;0,1;\)
a) Khi a = 6,5 thì :
\(\sqrt{a^2}\)
= \(\left|a\right|\)
=\(\left|6,5\right|\)
= 6,5
Khi a = -0,1 thì :
\(\sqrt{a^2}\)
= \(\left|a\right|\)
= \(\left|-0,1\right|\)
= \(-\left(-0,1\right)\)
= 0,1
Tuong tu nhu tren :
b) \(\sqrt{a^4}=\sqrt{\left(a^2\right)^2}\)
c)\(\sqrt{a^6}=\sqrt{\left(a^3\right)^2}\)
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bài 1: rút gọn các biểu thức.
a) dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-(sqrt{x}-sqrt{y})^2
b) sqrt{dfrac{x-2sqrt{x}+1}{x+2sqrt{x}+1}}(xge0)
c) dfrac{x-1}{sqrt{y}-1}sqrt{dfrac{(y-2sqrt{y}+1)^2}{(x-1)^4}}(xne1,yne1,y0)
bài 2:rút gọn và tính.
a) sqrt{dfrac{sqrt{a}-1}{sqrt{b}+1}:}sqrt{dfrac{sqrt{b}-1}{sqrt{a}+1}với}a7,25;b3,25
b) sqrt{15a^2-8asqrt{15}+16}vớiasqrt{dfrac{3}{5}}+sqrt{dfrac{5}{3}}
c) sqrt{10a^2-4asqrt{10}+4}vớiasqrt{dfrac{2}{5}}+sqrt{dfrac{5}{2}}
d) sqrt{a^2+2sqrt{a^2-1}}-sq...
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bài 1: rút gọn các biểu thức.
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})^2\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}(x\ge0)\)
c) \(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{(y-2\sqrt{y}+1)^2}{(x-1)^4}}(x\ne1,y\ne1,y>0)\)
bài 2:rút gọn và tính.
a) \(\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}:}\sqrt{\dfrac{\sqrt{b}-1}{\sqrt{a}+1}với}a=7,25;b=3,25\)
b) \(\sqrt{15a^2-8a\sqrt{15}+16}vớia=\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\)
c) \(\sqrt{10a^2-4a\sqrt{10}+4}vớia=\sqrt{\dfrac{2}{5}}+\sqrt{\dfrac{5}{2}}\)
d) \(\sqrt{a^2+2\sqrt{a^2-1}}-\sqrt{a^2-2\sqrt{a^2-1}}(a=\sqrt{5})\)
bài 3: rút gọn các biểu thức.
a) \(\sqrt{9(x-5)^2}(x\ge5)\)
b) \(\sqrt{x^2.(x-2)^2}(x< 0)\)
c)\(\dfrac{\sqrt{108x^3}}{\sqrt{12x}}(x>0)\)
d)\(\dfrac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}(x< 0:y\ne0)\)
ai giúp mik vs ạ, cảm ơn !
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
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