Bài 4: Liên hệ giữa phép chia và phép khai phương

\(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-394\right)\left(457+394\right)}}=\sqrt{\dfrac{73\cdot225}{63\cdot851}}\)

\(=\dfrac{2a+3b}{\sqrt{2a}+\sqrt{3b}}+\dfrac{\left(\sqrt{2a}-\sqrt{3b}\right)\left(\sqrt{4a^2}+\sqrt{2a\cdot3b}+\sqrt{9b^2}\right)}{3b-2a}\)

\(=\dfrac{\left(2a+3b\right)\left(\sqrt{2a}-\sqrt{3b}\right)-\left(\sqrt{2a}-\sqrt{3b}\right)\left(2a+\sqrt{6ab}+3b\right)}{2a-3b}\)

\(=\dfrac{\left(\sqrt{2a}-\sqrt{3b}\right)\cdot\left(-\sqrt{6ab}\right)}{2a-3b}=\dfrac{-\sqrt{6ab}}{\sqrt{2a}+\sqrt{3b}}\)

a: \(\Leftrightarrow\sqrt{\dfrac{x^2+2x-3}{x-1}}=x+3\)

\(\Leftrightarrow\sqrt{x+3}=x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(x+3\right)^2=x+3\end{matrix}\right.\Leftrightarrow x\in\left\{-3;-2\right\}\)

b: \(\Leftrightarrow\sqrt{x^2+2x+4}=\sqrt{19}\)

\(\Leftrightarrow x^2+2x+4-19=0\)

\(\Leftrightarrow x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>x=3

`a, 1/(mn^2) . sqrt ((m^2n^4)/5)

`= 1/5`.

`b, sqrt( m^4/(9 - 12 m + 4m^2))`

`= sqrt ((m^2)^2)/(3-2m)^2))`

`= m^2/(3-2m)`

`c, = (a-1)/(sqrt a-1) xx sqrt ((sqrt a -1)/(a-1)^4))`

`= sqrt(1/(a-1)^2)`

`= 1/|a-1|`

`d, (a-b)/(sqrt (a+b)) : sqrt ((a-b)^2/(a(a+b))`

`= (a-b)/(sqrt (a+b)) xx sqrt( (a(a+b))/(a-b)^2)`

`= sqrt (a/(a+b))`.

\(1,=\dfrac{13}{15}\\ 3,=\sqrt{\dfrac{3600}{121}}=\dfrac{60}{11}\\ 3,=\dfrac{3\left(a-4\right)}{4}\\ 4,=\dfrac{2\sqrt{3}}{14\sqrt{3}}=\dfrac{2}{14}=\dfrac{1}{7}\\ 5,=\dfrac{15}{\sqrt{84375}}=\dfrac{15}{75\sqrt{15}}=\dfrac{\sqrt{15}}{75}\\ 6,=2\sqrt{18}:\sqrt{2}-3\sqrt{8}:\sqrt{2}=6-6=0\)

1: =13/15

2: \(=\sqrt{\dfrac{13\cdot117}{121}}=\sqrt{\dfrac{13^2\cdot3^2}{11^2}}=\dfrac{39}{11}\)

3: \(=\sqrt{\dfrac{12}{588}}=\dfrac{1}{7}\)

4: \(=\sqrt{\dfrac{15^5}{5^5\cdot3^3}}=\sqrt{\dfrac{3^5}{3^3}}=3\)

Sửa đề: 2+căn 3

\(A=\dfrac{3+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}=\dfrac{\sqrt{6}\left(\sqrt{3}+1\right)}{2+\sqrt{3}+1}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}=\sqrt{2}\)

a: ĐK của A là x>3

ĐKXĐ của B là \(\dfrac{2x+3}{x-3}>=0\)

=>x>3 hoặc x<=-3/2

b: Để A=1/căn 2 thì \(\dfrac{2x^2+3}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow4x^2+6-x+3=0\)

\(\Leftrightarrow4x^2-x+9=0\)

hay \(x\in\varnothing\)

c: Để B=1/căn 2 thì \(\dfrac{2x+3}{x-3}=\dfrac{1}{2}\)

=>4x+6=x-3

=>3x=-9

hay x=-3(nhận)

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

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

b: \(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x+y\right)}{2}=\dfrac{\sqrt{3}}{x-y}\)

c: \(=\dfrac{x+\sqrt{7}}{\left(x+\sqrt{7}\right)^2}=\dfrac{1}{x+\sqrt{7}}\)

`a,sqrt((x-2 sqrt x+1)/(x+2sqrt x +1)`

` = sqrt((sqrt x-1)/(sqrt x+1)^2)`

`= (sqrt x- 1)/(sqrt x +1)`.

`b, = 2/((x-y)(x+y)) . ( sqrt. 3(x+y))/2`

`= sqrt 3/(x-y)`

`c, = (x+sqrt 7)/(x + sqrt 7)^2`

`= 1/(x + sqrt 7)^2`

`d, = (sqrt (xy) . (sqrt x + sqrt y))/(sqrt x + sqrt y)^2`

`= (sqrt(xy))/(sqrt x + sqrt y)`

Bài 4: 

a: =>|x-5|=7

=>x-5=7 hoặc x-5=-7

=>x=12 hoặc x=-2

b: \(\Leftrightarrow\dfrac{x-3}{2x+1}=4\)

=>x-3=8x+4

=>8x+4=x-3

=>7x=-7

hay x=-1(nhận)

c: \(\Leftrightarrow\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\)

=>5x-3=0 hoặc 5x+3=4

=>x=3/5 hoặc x=1/5

d: \(\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{5}\cdot3\sqrt{x-5}=3\)

\(\Leftrightarrow\sqrt{x-5}=3:\dfrac{12}{5}=\dfrac{15}{12}=\dfrac{5}{4}\)

=>x-5=25/16

=>x=105/16