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

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

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

`<=>2x-4=2\sqrt{4-x}`

`<=>x-2=\sqrt{4-x}`   với `2 <= x <= 4`

`<=>x^2-4x+4=4-x`

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

`<=>x(x-3)=0`

`<=>x=0` hoặc `x=3`

   (ko t/m)              (t/m)

Vậy `S={3}`

ĐK: \(-1\le x\le4\)

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

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

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

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

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

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

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

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

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

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

`\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}`

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

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

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

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

\(\Rightarrow1+x\sqrt{x^2+1}=\sqrt{x^2+1}+x\)

\(\Rightarrow1+x\sqrt{x^2+1}-\sqrt{x^2+1}-x=0\)

\(\Rightarrow-\left(x-1\right)+\left(x-1\right)\sqrt{x^2+1}=0\)

\(\Rightarrow\left(x-1\right)\left(\sqrt{x^2+1}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{x^2+1}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2+1}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+1=1\end{matrix}\right.\)

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

\(a,2y^2-x+2xy=y+4\\ \Leftrightarrow2y\left(x+y\right)-\left(x+y\right)=4\\ \Leftrightarrow\left(2y-1\right)\left(x+y\right)=4=4\cdot1=\left(-4\right)\left(-1\right)=\left(-2\right)\left(-2\right)=2\cdot2\)

Vì \(x,y\in Z\Leftrightarrow2y-1\) lẻ 

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

Vậy PT có nghiệm \(\left(x;y\right)=\left\{\left(3;1\right);\left(4;0\right)\right\}\)

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)