a, \(x^3+x^2+4\)

\(=x^3+2x^2-x^2-2x+2x+4=0\)

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

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

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

\(\left(x-1\right)^2=2.\left(x^2-1\right)\\ \Leftrightarrow x^2-2x+1=2x^2-2\\ \Leftrightarrow2x^2-x^2+2x-2-1=0\\ \Leftrightarrow x^2+2x-3=0\\ \Leftrightarrow x^2+3x-x-3=0\\ \Leftrightarrow x.\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x-1\right).\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ \Rightarrow S=\left\{-3;1\right\}\)

\(\Leftrightarrow2x+3\sqrt[3]{x^2-1}\left(\sqrt[3]{x-1}+\sqrt[3]{x+1}\right)=2x^3\)

\(\Rightarrow2x+3\sqrt[3]{x^2-1}.x\sqrt[3]{2}=2x^3\)

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

Xét (1):

Đặt \(\sqrt[3]{2x^2-2}=t\Rightarrow2x^2=t^3+2\)

\(\Rightarrow2+3t=t^3+2\)

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

\(\Leftrightarrow...\)

Lời giải:

a) Khi $m=2$ thì pt trở thành:

$x^2-10x+15=0\Leftrightarrow (x-5)^2=10\Rightarrow x=5\pm \sqrt{10}$
b) 

Để pt có 2 nghiệm pb $x_1,x_2$ thì trước tiên:

$\Delta'=(2m+1)^2-(4m^2-2m+3)>0$

$\Leftrightarrow 6m-2>0\Leftrightarrow m>\frac{1}{3}$

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2(2m+1)\\ x_1x_2=4m^2-2m+3\end{matrix}\right.\)

Để $(x_1-1)^2+(x_2-1)^2+2(x_1+x_2-x_1x_2)=18$

$\Leftrightarrow x_1^2+x_2^2-2(x_1+x_2)+2+2(x_1+x_2-x_1x_2)=18$

$\Leftrightarrow x_1^2+x_2^2-2x_1x_2=16$

$\Leftrightarrow (x_1+x_2)^2-4x_1x_2=16$

$\Leftrightarrow 4(2m+1)^2-4(4m^2-2m+3)=16$

$\Leftrightarrow (2m+1)^2-(4m^2-2m+3)=4$

$\Leftrightarrow 6m-2=4\Leftrightarrow m=1$ (thỏa mãn)

vậy...........

Lời giải:
ĐKXĐ: \(1\le x\leq 2\)

Ta có: \((\sqrt{2-x}+1)(\sqrt{x+3}-\sqrt{x-1})=4\)

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

\(\Leftrightarrow (\sqrt{2-x}+1).\frac{4}{\sqrt{x+3}+\sqrt{x-1}}=4\Rightarrow \sqrt{2-x}+1=\sqrt{x+3}+\sqrt{x-1}\)

\(\Leftrightarrow (\sqrt{x+3}-2)+\sqrt{x-1}-(\sqrt{2-x}-1)=0\)

\(\Leftrightarrow \frac{x-1}{\sqrt{x+3}+2}+\sqrt{x-1}-\frac{1-x}{\sqrt{2-x}+1}=0\)

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

Hiển nhiên biểu thức trong ngoặc lớn luôn lớn hơnm $0$

Do đó \(\sqrt{x-1}=0\Leftrightarrow x=1\) (thỏa mãn)

a.

\(\sqrt{x+4\sqrt{x}+4=5x+2}\)

\(\Rightarrow\sqrt{\left(\sqrt{x}\right)^2+2.2.\sqrt{x}+2^2}=5x+2\)

\(\Rightarrow\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

\(\Rightarrow\sqrt{x}+2=5x+2\)

\(\Rightarrow\sqrt{x}=5x\)

\(\Rightarrow x=25x^2\)

\(\Rightarrow x=0\)
Vậy nghiệm của phương trình là x = 0

b)

\(\sqrt{x-2\sqrt{x}+1}-\sqrt{x-4\sqrt{x}+4}=10\)

\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{\left(\sqrt{x}-2\right)^2=10}\)

\(\Rightarrow\sqrt{x}-1-\sqrt{x}+2=10\)

\(\Rightarrow1=10\) (Vô lí)

Vậy phương trình đã cho vô nghiệm

\(4+\dfrac{x}{5}=\dfrac{5}{3}\\ \Rightarrow\dfrac{x}{5}=-\dfrac{7}{3}\\ \Rightarrow x=-\dfrac{35}{3}\)

\(4+\dfrac{x}{5}=\dfrac{5}{3}\)

=>\(\dfrac{x}{5}=\dfrac{5}{3}-4=-\dfrac{7}{3}\)

=>x=-7/3*5=-35/3

x = -35/3