Bạn làm đc bài này chưa chỉ mình với

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: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)

\(=-4\sqrt{5a}+9\sqrt{a}\)

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) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)

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: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)

Bunyakovsky:

\(\sqrt{a+b}+\sqrt{a-b}\le\sqrt{2.2a}=2\sqrt{a}\)

a) \(\sqrt{a^3}\)-\(\sqrt{b^3}\)

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)(\(\sqrt{a}\)+\(\sqrt{b}\))

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).

\(\dfrac{\sqrt{45ab^2}}{\sqrt{20a}}=\sqrt{\dfrac{45ab^2}{20a}}=\sqrt{\dfrac{9}{4}b^2}=\dfrac{3}{2}b\)

\(\dfrac{\sqrt{45ab^2}}{\sqrt{20a}}=\sqrt{\dfrac{45ab^2}{20a}}=\sqrt{\dfrac{9b^2}{4}}=\dfrac{3b}{2}\)

a: \(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}-2b}{a-b}\)

\(=\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)

\(=\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(\left(a\right)\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)

\(\left(b\right) \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2\right)}{\sqrt{2}+\sqrt{3}+2}\\=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\\ =1+\sqrt{2}\)

\(\left(c\right)\sqrt{9\left(3-a\right)^2}vớia>3\\ =\sqrt{9}.\sqrt{\left(3-a\right)^2}\\ =3.\left|3-a\right|\\ =-3\left(3-a\right)vì.a>3\\ =3a-9\)

\(\left(d\right)\sqrt{a^2.\left(a-2\right)^2}vớia< 0\\ =\sqrt{\left[a\left(a-2\right)\right]^2}\\ =\left|a\left(a-2\right)\right|=-a.\left[-\left(a-2\right)\right]=a\left(a-2\right)=a^2-2a\)

Chúc bạn học tốt !

a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\ge2\sqrt{\sqrt{a^2+2}.\dfrac{1}{\sqrt{a^2+2}}}=2\)

Dấu = xảy ra khi \(\sqrt{a^2+2}=\dfrac{1}{\sqrt{a^2+2}}\Leftrightarrow a^2=-1\left(vn\right)\)

\(\Rightarrow\) Dấu "=" không xảy ra

Vậy \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

b)Với x,y>0,ta cm bđt phụ sau:

\(x^3+y^3\ge xy\left(x+y\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)

\(\Leftrightarrow\cdot\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) có:

\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}.\sqrt{b}}\ge\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" xra khi a=b

Vậy...