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

Đặt \(\sqrt{x^2+1}=t>0\)

\(\Rightarrow t^2-\left(x+3\right)t+3x=0\)

\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3+x-3}{2}=x\\t=\dfrac{x+3-x+3}{2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(x\ge0\right)\\\sqrt{x^2+1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(vô-nghiệm\right)\\x=\pm2\sqrt{2}\end{matrix}\right.\)

ĐK: Với mọi x thuộc R.

Ta có: \(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

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

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=\left(x+3\right)^2\left(x^2+1\right)\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=x^4+6x^3+10x^2+6x+9\)

\(\Leftrightarrow x^2-8=0\)

\(\Leftrightarrow x^2=8\)

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

Vậy....

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

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

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

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

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

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

Đặt \(\sqrt{x^2+x+1}=t>0\Rightarrow x^2=t^2-x-1\)

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

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

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

\(\Leftrightarrow\left(t+1\right)\left(t+x-3\right)=0\)

\(\Leftrightarrow t=3-x\)

\(\Leftrightarrow\sqrt{x^2+x+1}=3-x\) (\(x\le3\))

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

\(\Leftrightarrow x=\dfrac{8}{7}\)

a.

\(\Leftrightarrow2x^2\ge3\Leftrightarrow x^2\ge\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

b.

\(\Leftrightarrow\left(1-x\right)\left(x-3\right)\ge0\Rightarrow1\le x\le3\)

c.

\(\Leftrightarrow\sqrt{1-3x}\le2-x\Leftrightarrow\left\{{}\begin{matrix}1-3x\ge0\\2-x\ge0\\1-3x\le x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\le2\\x^2-x+3\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Delta=9-4m>0\Rightarrow m< \dfrac{9}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=m\end{matrix}\right.\)

\(\sqrt{x_1^2+1}+\sqrt{x_2^2+1}=3\sqrt{3}\)

\(\Leftrightarrow x_1^2+x_2^2+2+2\sqrt{\left(x_1^2+1\right)\left(x_2^2+1\right)}=27\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\sqrt{\left(x_1x_2\right)^2+\left(x_1+x_2\right)^2-2x_1x_2+1}=25\)

\(\Leftrightarrow9-2m+2\sqrt{m^2+9-2m+1}=25\)

\(\Leftrightarrow\sqrt{m^2-2m+10}=m+8\left(m\ge-8\right)\)

\(\Leftrightarrow m^2-2m+10=m^2+16m+64\)

\(\Rightarrow m=-3\) (thỏa mãn)

Pt trên có a=1, b=5, c=-3m+2

\(\Delta=b^2-4ac=25-4\cdot1\cdot\left(-3m+2\right)=17+12m\)

Để pt có hai nghiệm phân biệt thì \(\Delta>0\)<=> 17+12m >0  <=>m> 17/12

Theo hệ thức Viet, ta có:

\(\hept{\begin{cases}x_1+x_2=-5\\x_1\cdot x_2=-3m+2\end{cases}}\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1\cdot x_2=25-4\left(-3m+2\right)=17+12m=10\)

=> 12m = -7      <=>m=-7/12 (thỏa đkxđ)

Vậy với m=-7/12 thì phương trình có hai nghiệm x1, x2 thỏa (x1 - x2)^2 =10

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Giúp mk vs

a: \(=x^3-2x^5\)

e: \(=x^4+2x^3-x^2-2x\)

a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)

a,

\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b,

\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)