1) \(\sqrt[3]{x+1}=5\)

\(\Rightarrow x+1=125\)

\(\Rightarrow x=124\)

2) \(\sqrt[3]{1-3x^3}=-2\)

\(\Rightarrow1-3x^3=-8\)

\(\Rightarrow3x^3=9\)

\(\Rightarrow x=\sqrt[3]{3}\)

a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)

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

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

b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)

=>\(x-2\sqrt{x}-1=0\)

=>\(\Leftrightarrow x=3+2\sqrt{2}\)

c: |2x-5|=3

=>2x-5=3 hoặc 2x-5=-3

=>2x=2 hoặc 2x=8

=>x=4(nhận) hoặc x=1(loại)

Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)

1) Ta có: \(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1-21+x=0\)

\(\Leftrightarrow x^2-3x-20=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)

1)\(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow21-x=x^2-2x+1\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

2)\(\sqrt{8-x}+2=x\)

\(\Leftrightarrow8-x=\left(x-2\right)^2\)

\(\Leftrightarrow8-x=x^2-4x+4\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

3)\(1+\sqrt{3x+1}=3x\)

\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\sqrt{4-x^2}=\sqrt{x+2}\) (ĐK: \(-2\le x\le2\))

\(\Leftrightarrow4-x^2=x+2\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

_______

\(\sqrt{9x^2-4}=2\sqrt{3x-2}\) (ĐK: \(x\ge\dfrac{2}{3}\)) 

\(\Leftrightarrow9x^2-4=4\left(3x-2\right)\)

\(\Leftrightarrow9x^2-4=12x-8\)

\(\Leftrightarrow9x^2-12x+4=0\)

\(\Leftrightarrow\left(3x-2\right)^2=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

\(P=\frac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+\sqrt{x}-2}\)

\(P=\frac{3x+3\sqrt{x}-3-x+1-x+4}{x+\sqrt{x}-2}\)

\(P=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)=\left(\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right)-\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)\)

\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}=\frac{3\left(x^2-x-1\right)-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}-\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}\)

\(\Leftrightarrow\frac{3x^2-5x-2}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-4}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-6}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-1\right)\left(x-2\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)

Dễ thấy: \(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

sao cái từ "dễ thấy" nó khó thấy quá v 

haha :v

+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)    \(\left(ĐK:x\ge0\right)\)

        \(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)

        \(\Leftrightarrow3\sqrt{3x}=6\)

        \(\Leftrightarrow\sqrt{3x}=2\)

        \(\Leftrightarrow3x=4\)

        \(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{4}{3}\right\}\)

+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\)    \(\left(ĐK:x\ge1\right)\)

        \(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

        \(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)

        \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)

 Vậy \(S=\left\{1,15\right\}\)

+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\)       \(\left(ĐK:x\ge0\right)\)

         \(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

         \(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

   Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)

    \(\Rightarrow\)\(\sqrt{x}-4< 0\)

   \(\Leftrightarrow\)\(\sqrt{x}< 4\)

   \(\Leftrightarrow\)\(x< 16\)

   Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)

 Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)

\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)  (Đk: x \(\ge\)0)

<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)

<=> \(3\sqrt{3x}=6\)

<=> \(\sqrt{3x}=2\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)

<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\) 

<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)

\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)

<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

<=>  \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)

<=> \(\sqrt{x}< 4\) <=> \(x< 16\)

Kết hợp với đk => S = {x|0 < x < 16}

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

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

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290