\(x\left(x+2\right)\left(x^2+2x+2\right)+1=0\Leftrightarrow\left(x+1-1\right)\left(x+1+1\right)\left(x^2+2x+1+1\right)+1=0\) \(Đạt:x+1=a\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)+1=0\Leftrightarrow\left(a^2-1\right)\left(a^2+1\right)+1=0\Leftrightarrow a^4-1+1=0\Leftrightarrow a^4=0\Leftrightarrow a=0\Leftrightarrow x=-1.Vậy:x=-1\)

Bài làm:

Ta có: \(y^2+4^x+2y-2^{x+1}+2=0\)

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

\(\Leftrightarrow\left(y+1\right)^2+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)

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

Mà \(\hept{\begin{cases}\left(y+1\right)^2\ge0\\\left(2^x-1\right)^2\ge0\end{cases}}\forall x,y\)

\(\Rightarrow\left(y+1\right)^2+\left(2^x-1\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(y+1\right)^2=0\\\left(2^x-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-1\\2^x=1=2^0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy \(\left(x;y\right)=\left(0;-1\right)\)

Cảm ơn bạn nhiều nha !

bài 2:

c)    \(x^3+8x^2+17x+10=0\)

\(\Leftrightarrow\)\(x^3+x^2+7x^2+7x+10x+10=0\)

\(\Leftrightarrow\)\(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2+7x+10\right)=0\)

đến đây thì dễ rồi, bn cm  x^2 + 7x + 10 > 0 

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

\(\Leftrightarrow x^2=-2 \) (vô lý)

Vậy \(S=\left\{\varnothing\right\}\)

Bài 2:

a) Khi m = - 2, phương trình (1) trở thành:\(x^2-6x-7=0\)

\(\Delta=b^2-4ac=\left(-6^2\right)-4.\left(-7\right)=64\)

\(\sqrt{\Delta}=\sqrt{64}=8>0\)

Phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{6+8}{2}=7\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{6-8}{2}=-1\)

Vậy \(S=\left\{7;-1\right\}\)

\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (ĐKXĐ: \(x\ne0;x\ne\dfrac{1}{2}\))

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

\(\Leftrightarrow2x^2+6x-x-3=2x^2+2x\)

\(\Leftrightarrow2x^2-2x^2+6x-x-2x=3\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)\)

\(\Rightarrow S=\left\{1\right\}\)

\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=x\left(2x+2\right)\)

\(\Leftrightarrow2x^2-x+6x-3=2x^2+2x\)

\(\Leftrightarrow2x^2+5x-3-2x^2-2x=0\)

\(\Leftrightarrow3x-3=0\)

\(\Leftrightarrow3\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

Đặt \(\sqrt{6x^2-12x+7}=t\left(t\ge0\right)\)

<=>\(t^2-7=6x^2-12x\)

\(\Leftrightarrow\dfrac{t^2-7}{6}=x^2-2x\)

Ta có pt mới:

\(\dfrac{7-t^2}{6}+t=0\)

\(\Leftrightarrow t^2-6t-7=0\)

\(\Leftrightarrow t^2-2\cdot t\cdot3+9-9-7=0\)

\(\Leftrightarrow\left(t-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}t=7\\t=-1\end{matrix}\right.\)(loại t=-1)

Với t=7

=>\(\sqrt{6x^2-12x+7}=7\)

<=>6x²-12x+7=49

<=>6x²-12x-42=0

<=>x²-2x-7=0

<=>(x-1)²=8

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

\(\Leftrightarrow x^2+4=2x+3\)

=>x^2-2x+1=0

=>(x-1)^2=0

=>x=1