Bài 9: Căn bậc ba

`C_2:`

`a)\sqrt{x^2+6x+9}=5`         

`<=>\sqrt{(x+3)^2}=5`

`<=>|x+3|=5`

`<=>[(x+3=5),(x+3=-5):}<=>[(x=2),(x=-8):}`

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Có vẻ đề phải là: `\sqrt{9x-9}?`

`b)\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8`      `ĐK: x >= 1`

`<=>4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8`

`<=>4\sqrt{x-1}=8`

`<=>\sqrt{x-1}=2`

`<=>x-1=4<=>x=5` (t/m)

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`C_3:`

`a)A` xác định `<=>{(x > 0),(\sqrt{x}-1 \ne 0),(\sqrt{x}-2 \ne 0):}<=>{(x > 0),(x \ne 1,x \ne 2):}`

Với `x > 0,x \ne 1,x \ne 2` có:

`A=(1/[\sqrt{x}-1]-1/\sqrt{x}):([\sqrt{x}+1]/[\sqrt{x}-2]-[\sqrt{x}+2]/[\sqrt{x}-1])`

`A=[\sqrt{x}-\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]:[(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)]/[(\sqrt{x}+2)(\sqrt{x}-1)]`

`A=1/[\sqrt{x}(\sqrt{x}-1)].[(\sqrt{x}+2)(\sqrt{x}-1)]/[x-1-x+4]`

`A=1/\sqrt{x} .[\sqrt{x}+2]/3`

`A=[\sqrt{x}+2]/[3\sqrt{x}]`

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`b)` Với `x > 0,x \ne 1,x \ne 2` có:

`A=1/4<=>[\sqrt{x}+2]/[3\sqrt{x}]=1/4`

          `<=>4\sqrt{x}+8=3\sqrt{x}`

         `<=>\sqrt{x}=-8` (Vô lí)

 `=>` Ko có gtr của `x` t/m

`1)` Căn thức có nghĩa `<=>1/2x-3 >= 0`

                                     `<=>1/2x >= 3<=>x >= 6`

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`2)`

`a)\sqrt{12}+\sqrt{48}-\sqrt{27}+3\sqrt{47}`

`=2\sqrt{3}+4\sqrt{3}-3\sqrt{3}+3\sqrt{47}`

`=3\sqrt{3}+3\sqrt{47}`

~~~~~~~~~~~~

`b)3/2\sqrt{6}+\sqrt{2/3}-5\sqrt{3/2}`

`=3/2\sqrt{6}+1/3\sqrt{6}-5/2\sqrt{6}`

`=-2/3\sqrt{6}`

~~~~~~~~~~~~

`c)\root[3]{-125}+\root[3]{343}-\root[3]{64}`

`=-5+7-4`

`=-2`

\(D=\dfrac{1-x-\sqrt{x^3}}{\sqrt{x}+1}\) \(\left(x\ge0\right)\)

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

\(D=\dfrac{1-x\left(1+\sqrt{x}\right)}{\sqrt{x}+1}\)

\(D=-\dfrac{x\left(1+\sqrt{x}\right)}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}\)

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

a: \(2\sqrt{5}-5\sqrt{2}< 0< 1\)

b: \(\sqrt{\dfrac{8}{3}}=\dfrac{2\sqrt{2}}{\sqrt{3}}=\dfrac{2\sqrt{6}}{3}< \dfrac{3}{4}\)

\(x=\sqrt[3]{16-\sqrt{255}}+\sqrt[3]{16+\sqrt{255}}\)

\(\Leftrightarrow x^3=16+16+3\cdot x\)

=>x³-3x-32=0

=>\(x\simeq3.49\)

⇔ $x^{2}.(x+2)=(x+2)^{3}$

⇔$4(x+1)(x+2)=0$

→$x=-1 hoặc x=-2$

P/t \(\Leftrightarrow\sqrt[3]{x^2\left(x+2\right)}=x+2\)  

\(\Leftrightarrow\left(x+2\right)\left(\sqrt[3]{x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm1\end{matrix}\right.\)

Lời giải:
ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow x^3+2x^2=(x+2)^3$

$\Leftrightarrow x^2(x+2)-(x+2)^3=0$

$\Leftrightarrow (x+2)[x^2-(x+2)^2]=0$

$\Leftrightarrow (x+2)(x-x-2)(x+x+2)=0$

$\Leftrightarrow (x+2)(-2).2(x+1)=0$

$\Leftrightarrow (x+2)(x+1)=0$

$\Leftrightarrow x+2=0$ hoặc $x+1=0$

$\Leftrightarrow x=-2$ hoặc $x=-1$

a: \(=\left(\sqrt[3]{5}\right)^3+\left(\sqrt[3]{2}\right)^3=5+2=7\)

b: \(=\sqrt[3]{\left(\sqrt[3]{3}-1\right)^3}=\sqrt[3]{3}-1\)

\(f=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\right)}{\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}}=\sqrt[3]{2}\)

e: \(=5+3+3\sqrt[3]{75}+\sqrt[3]{45}-3\sqrt[3]{75}-3\sqrt[3]{45}=8\)

e: =>x+4+1-x-2căn (x+4)(1-x)=1-2x

=>2căn (x+4)(1-x)=5-1+2x=2x+4

=>căn (x+4)(1-x)=x+2

=>(x+4)(1-x)=x^2+4x+4 và x>=-2

=>x^2+4x+4=x-x^2+4-4x và x>=-2

=>2x^2+7x=0 và x>=-2

=>x=0

f: Đặt \(\sqrt[3]{x-1}=a;\sqrt[3]{x-2}=b\)

=>a^3+b^3=2x-3

\(\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2x-3}\)

=>a^3+b^3=a^3+b^3+3ba(a+b)

=>ab(a+b)=0

=>x-1=0 hoặc x-2=0 hoặc x-1=-x+2

=>x=1;x=2;x=3/2