\(b,\left(x^2+1\right)^2+3x\left(X^2+1\right)+2x^2=0\)

đặt x^2+1 là y ta đc

\(y^2+3xy+2x^2=0< =>y^2+2xy+xy+2x^2=0< =>y\left(y+2x\right)+x\left(y+2x\right)=0< =>\left(y+x\right)\left(y+2x\right)=0< =>\left[{}\begin{matrix}y=-x\left(1\right)\\y=-2x\left(2\right)\end{matrix}\right.\)

giải 1 ta có;\(x^2+1=-x< =>x^2+x+1=0< =>x^2+2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0< =>\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\left(vônghiemej\right)\)

giải 2:\(x^2+1=-2x< =>x^2+2x+1=0< =>\left(x+1\right)^2=0< =>x+1=0< =>x=-1\left(nhận\right)\)

vậy......

b)Cách khác:\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

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

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

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

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

\(\Leftrightarrow x=-1\)

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

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

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

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

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

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

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

pt <=> \(\left(3x+1+x-7\right)\left(\left(3x+1\right)^2+\left(x-7\right)^2\right)=\left(4x-6\right)^3\)

\(\Leftrightarrow\left(4x-6\right)\left(9x^2+6x+1+x^2-14x+49-\left(4x-6\right)^2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(10x^2-8x+50-16x^2+48x-36\right)=0\)

\(\orbr{\begin{cases}2x-3=0\\-6x^2+40x+14=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\-3x^2+20x+7=0\left(\cdot\right)\end{cases}}\)

pt(*) <=> (3x-1)(x+7)=0 <=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-7\end{cases}}\)

Vậy x=...

${}^{}+{}^{}={}^{}\iff {}^{}+3{}^{}+3x+1+{}^{}-6{}^{}+12x-8=8{}^{}-12{}^{}+6x-1\iff 2{}^{}-3{}^{}+15x-7-8{}^{}+12{}^{}-6x+1=0$$\iff -6{}^{}$

Đặt x+1=a; x-2=b

Phương trình trở thành:

\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;2;\dfrac{1}{2}\right\}\)

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\)

\(\Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0\)

\(\Leftrightarrow-6x^3+9x^2+9x-6=0\)

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

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

\(\Leftrightarrow2\left(x^3+1\right)-3x\left(x+1\right)=0\)

\(\Leftrightarrow2\left(x^2-x+1\right)\left(x+1\right)-3x\left(x+1\right)=0\)

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

\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[\left(2x^2-4x\right)-\left(x-2\right)\right]\left(x+1\right)=0\)

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

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

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(\text{Vậy tập nghiệm phương trình là:}\left\{\dfrac{1}{2};2;\left(-1\right)\right\}\)

c.

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

Đặt \(\sqrt{x^2+3}=t>0\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

a.

Đề bài ko chính xác, pt này ko giải được

b.

ĐKXĐ: \(x\ge-\dfrac{7}{2}\)

\(2x+7-\left(2x+7\right)\sqrt{2x+7}+x^2+7x=0\)

Đặt \(\sqrt{2x+7}=t\ge0\)

\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)

\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)

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

a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).