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 (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

Bình phương 2 vế lên ta được:

\(x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}=16\) 16

Kết hợp bài ra ta thu được x=12

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

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

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

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

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

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

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

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

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

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

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

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

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 \(\left(x;y\right)=\left\{\left(3;1\right);\left(4;0\right)\right\}\)

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 \(x=8\)

hello bạn

ĐK: \(-1\le x\le4\)

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow2x^2+2+2\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=2x^2+4\)

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

\(\Leftrightarrow x^4+x^2=0\)

\(\Leftrightarrow x=0\)

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

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

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

`<=>x^4+x^2=0`

`<=>x=0`