Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)

a, đk x > 0 ; x khác 1 

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

\(=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{-2\sqrt{x}+1}{x\sqrt{x}-1}:\dfrac{-2\sqrt{x}+1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x}{x+\sqrt{x}+1}\)

b, Ta có \(K=\dfrac{x}{x+\sqrt{x}+1}=\dfrac{1}{3}\Leftrightarrow3x=x+\sqrt{x}+1\Leftrightarrow2x-\sqrt{x}-1=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)=0\Leftrightarrow x=1\)

c, bạn tự làm nhé 

\(3,=\sqrt{25.2}-\dfrac{\sqrt{9}}{\sqrt{2}}+3\sqrt{4.6}\\ =5\sqrt{2}-\dfrac{3}{\sqrt{2}}+6\sqrt{6}\\ =\dfrac{5\sqrt{2}.2-3\sqrt{2}+6\sqrt{6}.2}{2}\\ =\dfrac{10\sqrt{2}-3\sqrt{2}+12\sqrt{6}}{2}\\ =\dfrac{12\sqrt{6}+7\sqrt{2}}{2}\)

Giúp mik vs

`a)1/[1+\sqrt{2}]=[1-\sqrt{2}]/[1-2]=[1-\sqrt{2}]/[-1]=\sqrt{2}-1`

`b)\sqrt{3}/[\sqrt{3}-5]=[\sqrt{3}(\sqrt{3}+5)]/[3-25]=[-3-5\sqrt{3}]/22`

`c)[4\sqrt{5}]/[\sqrt{5}-\sqrt{3}]=[4\sqrt{5}(\sqrt{5}+\sqrt{3})]/[5-3]=2\sqrt{5}(\sqrt{5}+\sqrt{3})=10+2\sqrt{15}`

`d)[3\sqrt{3}-2\sqrt{2}]/[3\sqrt{3}+2\sqrt{2}]`

`=[(3\sqrt{3}-2\sqrt{2})^2]/[27-8]`

`=[27+8-12\sqrt{6}]/19`

`=[35-12\sqrt{6}]/19`

`e)[8\sqrt{3}-3\sqrt{5}]/[9\sqrt{3}-4\sqrt{5}]`

`=[(8\sqrt{3}-3\sqrt{5})(9\sqrt{3}+4\sqrt{5})]/[243-80]`

`=[216+32\sqrt{15}-27\sqrt{15}-60]/163`

`=[156+5\sqrt{15}]/163`

\(a,=\dfrac{9\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}\\ =\dfrac{3\left(3\sqrt{2}+\sqrt{3}\right)}{5}\\ =\dfrac{9\sqrt{2}+3\sqrt{3}}{5}\\ b,???\)

Bài 55:

a: \(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

b: \(=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)

lxxxxxxxxxx

Lx bn ey

a: \(=\dfrac{\sqrt{5b}}{7a\sqrt{a}}\)

b: \(=7xy\cdot\dfrac{\sqrt{-3x}}{\sqrt{xy}}=7\sqrt{xy\cdot\left(-3x\right)}\)

c: \(=\sqrt{\dfrac{5}{59}}\cdot\dfrac{x\sqrt{x}}{\sqrt{y}}\cdot x=\sqrt{\dfrac{5}{59}}\cdot\dfrac{x^2\sqrt{x}}{\sqrt{y}}\)

a: \(=2+\sqrt{3}-2\sqrt{3}-3=-1-\sqrt{3}\)

b: \(=3\sqrt{2}-2\sqrt{3}-4\sqrt{3}+\sqrt{2}=4\sqrt{2}-6\sqrt{3}\)

`a, = 2 - 2 sqrt 3 + sqrt 3 - 3`

`= -1 - sqrt 3`

`b, = (12 - 2 sqrt 12 . sqrt 18 + 18) - (48 - 2 . sqrt 48 . sqrt 2 + 2)`

`= (sqrt(sqrt 12 - sqrt 18))^2 - (sqrt(sqrt 48 - sqrt 2))^2`

`= sqrt 18 - sqrt 12 - sqrt 48 + sqrt 2`

`= -6 sqrt 3 + 4 sqrt 2`

\(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)