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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: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
\(\left(a\right)\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
\(\left(b\right) \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2\right)}{\sqrt{2}+\sqrt{3}+2}\\=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\\ =1+\sqrt{2}\)
\(\left(c\right)\sqrt{9\left(3-a\right)^2}vớia>3\\ =\sqrt{9}.\sqrt{\left(3-a\right)^2}\\ =3.\left|3-a\right|\\ =-3\left(3-a\right)vì.a>3\\ =3a-9\)
\(\left(d\right)\sqrt{a^2.\left(a-2\right)^2}vớia< 0\\ =\sqrt{\left[a\left(a-2\right)\right]^2}\\ =\left|a\left(a-2\right)\right|=-a.\left[-\left(a-2\right)\right]=a\left(a-2\right)=a^2-2a\)
Chúc bạn học tốt ! 
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 !!!
a) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)
b)\(\sqrt{9\left(3-a\right)^2}=3\left|3-a\right|=3\left(a-3\right)\)(vì a > 3)
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(\sqrt{9}\right)^2}-\sqrt{\left(\sqrt{17}\right)^2}\)
\(\sqrt{9\left(3-a\right)^2}\)
\(=\sqrt{3^2\left(3-a\right)^2}\)
\(=3\left(3-a\right)\)
\(=3-3a\)
tách 11 ra thành \(\sqrt{3}\) mũ 2 + căn 8 mũ 2
áp dụng hẳng đẳng thức đáng nhớ A^2+2AB +B^2=(A+B)^2
vào \(\sqrt{11+4\sqrt{6}}\)
.Bản thử đi nhé kết quả của mình là \(\sqrt{3}\)+\(\sqrt{8}\)
Vì ko gõ đc căn nên mình ko giải hẳn hoi ra đc .Bạn thông cảm ha.
Chúc bn hok tốt!
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).
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)(\(\sqrt{a}\)+\(\sqrt{b}\))
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)\(\sqrt{a^6}=\sqrt{\left(a^3\right)^2}\)