a)\(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)

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

thế này à

a) \(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}=2\sqrt{196}-10\sqrt{245}+14\sqrt{784}\)

\(=28-10\sqrt{49.5}+392=420-70\sqrt{5}\)

b) \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}=\sqrt{144}-3\sqrt{48}+5\sqrt{16}+4\sqrt{9.3}\)

\(=12-3\sqrt{16.3}+20+12\sqrt{3}=32-12\sqrt{3}+12\sqrt{3}=32\)

a: \(\sqrt{\dfrac{3}{2}a^2}=\left|a\right|\cdot\dfrac{\sqrt{6}}{2}\)

b: \(\sqrt{\dfrac{1}{600}}=\dfrac{1}{10\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{6}{100}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{10}{196}}=\dfrac{1}{14}\cdot\sqrt{10}\)

c: \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)

d: căn 2/3=căn 6/9=1/3*căn 6

e: \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{25}}=\pm\dfrac{x\sqrt{5}}{5}\)

f: \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3x}{x^2}}=\dfrac{\sqrt{3x}}{\left|x\right|}\)

a) \(\sqrt{2x-3}=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\))

<=> \(\left\{{}\begin{matrix}x\ge3\\2x-3=\left(x-3\right)^2\left(1\right)\end{matrix}\right.\)

(1) <=> \(2x-3=x^2-6x+9\)

<=> \(x^2-8x+12=0\)

<=> (x-2)(x-6) = 0 <=> \(\left[{}\begin{matrix}x=2\left(l\right)\\x=6\left(c\right)\end{matrix}\right.\)

KL: Phương trình có nghiệm duy nhất x = 6
b) \(\sqrt{10-x}+\sqrt{x+3}=5\) (ĐK: \(-3\le x\le10\))

<=> \(\left(\sqrt{10-x}+\sqrt{x+3}\right)^2=25\)

<=> \(10-x+x+3+2\sqrt{\left(10-x\right)\left(x+3\right)}=25\)

<=> \(\sqrt{\left(10-x\right)\left(x+3\right)}=6\)

<=> (10-x)(x+3) = 36

<=> 7x - x² + 30 = 36

<=> x² -7x + 6 = 0

<=> (x-1)(x-6) = 0

<=> \(\left[{}\begin{matrix}x=1\left(c\right)\\x=6\left(c\right)\end{matrix}\right.\)

KL: Phương trình có nghiệm S = {1;6}

c) \(\sqrt{x+3}-\sqrt{x-4}=1\) (ĐK: \(x\ge4\))

<=> \(\sqrt{x+3}=\sqrt{x-4}+1\)

<=> \(x+3=x-4+1+2\sqrt{x-4}\)

<=> \(\sqrt{x-4}=3\)

<=> x-4 = 9 <=> x = 13 (c)

KL: Phương trình có nghiệm duy nhất x = 13

a) ĐK: `x≥3`

`\sqrt(2x-3)=x-3`

`<=>2x-3=(x-3)^2`

`<=>2x-3=x^2-6x+9`

`<=>x^2-8x+12=0`

`<=>` \(\left[{}\begin{matrix}x=6\left(TM\right)\\x=2\left(L\right)\end{matrix}\right.\)

Vậy `x=2`.

b) ĐK: `-3<=x<=10`

`\sqrt(10-x)+\sqrt(x-3)=5`

`<=>10-x+x-3+2\sqrt((10-x)(x-3))=25`

`<=>2\sqrt((10-x)(x-3))=18`

`<=>\sqrt((10-x)(x-3))=9`

`<=>(10-x)(x-3)=81`

`<=>-x^2+13x-30=81`

`<=>x^2-13x+111=0` (VN)

a) \(\sqrt{28+10\sqrt{3}}=\sqrt{\left(5+\sqrt{3}\right)^2}=\left|5+\sqrt{3}\right|=5+\sqrt{3}\)

b) \(\sqrt{7+3\sqrt{5}}=\sqrt{\left(\dfrac{3\sqrt{2}}{2}+\dfrac{\sqrt{10}}{2}\right)^2}=\dfrac{3\sqrt{2}}{2}+\dfrac{\sqrt{10}}{2}\)