Bài 9: Căn bậc ba

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

\(\Leftrightarrow\left[\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right].\dfrac{2}{\sqrt{x}-1}\)

\(\Leftrightarrow\left[\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right].\dfrac{2}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{x+1-2\sqrt{x}}{x\sqrt{x}-1}.\dfrac{2}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{x\sqrt{x}-1}.\dfrac{2}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)2}{x+\sqrt{x}+1}\)

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

a: =>x^3=x

=>x(x-1)(x+1)=0

=>\(x\in\left\{0;1;-1\right\}\)

b: =>x^3+9x^2=x^3+9x^2+27x+27

=>x=-1

c: =>5x^3-x^3+3x^2-3x+1=0

=>4x^3+3x^2-3x+1=0

=>x=-1,41

góc C=90-60=30 độ

Xét ΔABC vuông tại A có sin C=AB/BC

=>AB/BC=1/2

=>AB=5cm

=>AC=5 căn 3(cm)

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

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

`\sqrt{(1-\sqrt{7})^2}`

`=|1-\sqrt{7}|`

`=\sqrt{7}-1` (Vì `1 < \sqrt{7}`)

Sửa:`\sqrt{(4x-1)^2}=3`

`<=>|4x-1|=3`

`<=>[(4x-1=3),(4x-1=-3):}`

`<=>[(x=1),(x=-1/2):}`

Chịu chết

? bài đâu dạ

\(\sqrt[3]{-\dfrac{1}{8}}=\dfrac{-1}{2}\)

Căn bậc ba của \(\dfrac{-1}{8}\) = \(\dfrac{-1}{8}.\dfrac{-1}{8}.\dfrac{-1}{8}=\dfrac{-1}{512} =>C\)

ý B nha bạn