\(\sqrt{x-4\sqrt{x-1}+3}+\sqrt{x-6\sqrt{x-1}+8}=1\\ < =>\sqrt{x-1-2\sqrt{x-1}.2+4}+\sqrt{x-1-2\sqrt{x-1}.3+9}=1\\ < =>\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)ĐK: x>=1

\(< =>|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ < =>\left(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\right)^2=1\\ < =>\sqrt{x-1}-2+2\left|\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-3\right)\right|+\sqrt{x-1}-3=1\\ < =>2\sqrt{x-1}-5+2\left|x+5-5\sqrt{x-1}\right|=1\\ < =>2\left|x+5-5\sqrt{x-1}\right|=6-2\sqrt{x-1}\\ < =>\left|x+5-5\sqrt{x-1}\right|=3-\sqrt{x-1}\)

\(< =>\left[{}\begin{matrix}x+5-5\sqrt{x-1}=3-\sqrt{x-1}\left(1\right)\\x+5-5\sqrt{x-1}=\sqrt{x-1}-3\left(2\right)\end{matrix}\right.\)

Giải (1): \(x+5-5\sqrt{x-1}=3-\sqrt{x-1}\\ < =>x+2-4\sqrt{x-1}=0\\ < =>x-1-2\sqrt{x-1}.2+4=1\\ < =>\left(\sqrt{x-1}-2\right)^2=1\\ < =>\left[{}\begin{matrix}\sqrt{x-1}-2=1\\\sqrt{x-1}-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=0\left(loại\right)\end{matrix}\right.\)

Giải (2) cũng ra x=8

Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x^3-3x^2+6x-4=0\left(1\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

`\sqrt{14-x}-\sqrt{x-4}=\sqrt{x-1}`      `ĐK: 4 <= x <= 14`

`<=>\sqrt{14-x}=\sqrt{x-1}+\sqrt{x-4}`

`<=>14-x=x-1+x-4+2\sqrt{(x-1)(x-4)}`

`<=>14-x=2x-5+2\sqrt{x^2-5x+4}`

`<=>2\sqrt{x^2-5x+4}=19-3x`    với `4 <= x <= 19/3`

`<=>4(x^2-5x+4)=361-114x+9x^2`

`<=>4x^2-20x+16=361-114x+9x^2`

`<=>5x^2-94x+345=0`

`<=>5x^2-25x-69x+345=0`

`<=>(x-5)(5x-69)=0`

`<=>x=5` hoặc `x=69/5`

       (t/m)                    (ko t/m)

Vậy `S={5}`

\(\dfrac{-17}{15}\)

ĐK: \(-1\le x\le4\)

\(\sqrt{x+1}+\sqrt{4-x}=t\left(\sqrt{5}\le t\le\sqrt{10}\right)\Rightarrow\sqrt{-x^2+3x+4}=\dfrac{t^2-5}{2}\)

\(pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\)

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

\(\Leftrightarrow\left(t-3\right)\left(t+5\right)=0\)

\(\Leftrightarrow t=3\left(\text{Vì }\sqrt{5}\le t\le\sqrt{10}\right)\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}=3\)

\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)

\(\Leftrightarrow\sqrt{-x^2+3x+4}=2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

2\(\sqrt{x+2+\sqrt{x+1}}\) - \(\sqrt{x+1}\) = 4;  Đk \(x\ge\) -1

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

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

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

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

  \(\sqrt{x+1}\)       = 4 - 2

   \(\sqrt{x+1}\)       = 2

    \(x+1\)      = 4

    \(x\)              = 4 - 1

       \(x\)            = 3

\(...\Rightarrow2\sqrt[]{x+1+2\sqrt[]{x+1+1}}-\sqrt[]{x+1}=4\left(x\ge-1\right)\)

\(\Rightarrow2\sqrt[]{\left(\sqrt[]{x+1}+1\right)^2}-\sqrt[]{x+1}=4\)

\(\Rightarrow2|\sqrt[]{x+1}+1|-\sqrt[]{x+1}=4\left(1\right)\)

Nếu \(\sqrt[]{x+1}+1\ge0\Rightarrow x\ge-1\)

\(\left(1\right)\Rightarrow2\sqrt[]{x+1}+1-\sqrt[]{x+1}=4\)

\(\Rightarrow\sqrt[]{x+1}=3\Rightarrow x+1=9\Rightarrow x=8\)

Nếu \(\sqrt[]{x+1}+1\le0\Rightarrow x\in\varnothing\)

Vậy \(x=8\)