a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

<=>\(\left(2x^2+2\right)^2-\left(x^2-5x-2\right)^2=0\)

<=>\(\left(2x^2+2-x^2+5x+2\right)\left(2x^2+2+x^2-5x-2\right)=0\)

<=>\(\left(x^2+5x+4\right)\left(3x^2-5x\right)=0\)

<=>\(\left(x+1\right)\left(x+4\right)x\left(3x-5\right)=0\)

<=>x+1=0 hoặc x+4=0 hoặc x=0 hoặc 3x-5=0

<=>x=-1 hoặc x=-4 hoặc x=0 hoặc x=5/3

bài này dùng hằng đẳng thức a²-b²= (a-b)(a+b)

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

\(\left(x^2+5x+4\right)\left(3x^2-5x\right)=0\)

Mã mở

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1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

@Lightning Farron

@soyeon_Tiểubàng giải

@Trần Việt Linh

2) pt đề bài cho=0

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

<=>\(\orbr{\begin{cases}x-1=0\left(1\right)\\2x^2-x+2=0\left(2\right)\end{cases}}\)

Từ 1 => x=1

từ 2 =>\(2\left(x^2-\frac{1}{2}x+1\right)\)

 =\(2\left[\left(x-\frac{1}{4}\right)^2+\frac{15}{16}\right]>0\)với mọi x

Nên pt 2 cô nghiệm

Vậy pt đề cho có nghiệm là 1

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

3/ x(x + 3)(x + 1)(x + 2) = 24

   => (x² + 3x)(x² + 3x + 2) = 24

   Đặt a = x² + 3x ta được pt: a(a + 2) = 24 => a² + 2a - 24 = 0 => a = 4 hoặc a = -6

                                                         Vậy x = 1 , x = -4

4/ (x + 1)⁴ + (x + 3)⁴ = 2

    Đặt a = x + 2 ta được: (a - 1)⁴ + (a + 1)⁴ = 2

 \(\Rightarrow\left[\left(a-1\right)^2+\left(a+1\right)^2\right]^2-2\left(a-1\right)^2\left(a+1\right)^2=2\)

 \(\Rightarrow\left[\left(a-1+a+1\right)^2-2\left(a-1\right)\left(a+1\right)\right]^2-2\left(a^2-1\right)^2=0\)

 \(\Rightarrow\left[\left(2a\right)^2-2\left(a^2-1\right)\right]^2-2\left(a^2-1\right)^2=0\)

 \(\Rightarrow\left[4a^2-2\left(a^2-1\right)+\sqrt{2}\left(a^2-1\right)\right]\left[4a^2-2\left(a^2-1\right)-\sqrt{2}\left(a^2-1\right)\right]=0\)

\(\Rightarrow\left[\left(2+\sqrt{2}\right)a^2+2-\sqrt{2}\right]\left[\left(2-\sqrt{2}\right)a^2+2+\sqrt{2}\right]=0\)

                                          Tới đây bạn giải ra a rồi tính ra x nha

2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)

\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)

Đặt \(x^2+5x+3=t\)

\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)

\(\Leftrightarrow t^2-9=280\)

\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)

\(\Leftrightarrow x^2-2x+7x-14=0\)

\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0

\(\Leftrightarrow\) x = 2 hoặc x = - 7

Vậy x = 2 hoặc x = -7.

3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)

\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)

\(\Leftrightarrow x^3+12x^2+46x+60=0\)

\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)

\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)

\(\Leftrightarrow x=-6\)

Vậy x = -6.

4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow2\left[\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right]=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{6}{x\left(x+6\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2x\left(x+6\right)=54\)

\(\Leftrightarrow2x^2+12x-54=0\)

\(\Leftrightarrow2x^2-6x+18x-54=0\)

\(\Leftrightarrow2x\left(x-3\right)+18\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+18\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x+9\right)=0\)

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 9 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -9

Vậy x = 3 hoặc x = -9.

\(x^2+6x+6+\left(\frac{x+3}{x+2}\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)^2+\left(\frac{x+3}{x+4}\right)^2-3=0\)

đặt x+3=y => x+4=y+1

lại có \(y^2+\frac{y^2}{\left(y+1\right)^2}-3=0\)

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