\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)

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

\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)

\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)

trên là 12 thôi ạ

x^8 + 2x^6 + 2x^4 + x^2 + 1 - 4x^6 = 12( x^4 - 2x^2 - 1 ) - 4

x^8 + 2x^4 + x^2 + 1 - 2x^6 = 12x^4 - 24x^2 - 12 - 4

x^8 - 2x^6 = 12x^4 - 2x^4 - 24x^2 - x^2 - 16 - 1

x^8 - 2x^6 = 10x^4 - 25x^2 - 17

( x^2 )^4 - 2( x^2 )^3 = 10(x^2)^2 - 25x^2 - 17

0 = 10(x^2)^2 - ( x^2)^4 - 25x^2 + 2(x^2)^3 - 17

17 = (x^2)[ 10x^2 - (x^2)^3 - 25 + 2(x^2)^2 ]

17 = ( x^2 )[ 10x^2 - x^6 - 25 + 2x^4 ]

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Giải phương trình mà NEVER_NNL

Ta đặt \(x^2+x=a\)

 Khi đó pt trở thành :

\(a^2+4a=12\)

\(\Leftrightarrow a^2+4a-12=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=2\\a=-6\end{cases}}\)

Với \(a=2\Leftrightarrow x^2+x=2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Với \(a=-6\Leftrightarrow x^2+x=-6\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\) ( vô lí )

Vậy pt đã cho có tập nghiêm \(S=\left\{1,-2\right\}\)

Ta có: \(\Delta=4^2+4.12=64,\sqrt{\Delta}=8\)

\(\Rightarrow\orbr{\begin{cases}x^2+x=\frac{-4+8}{2}=2\\x^2+x=\frac{-4-8}{2}=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+x-2=0\\x^2+x+6=0\end{cases}}\)

+) \(x^2+x-2=0\)

Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\)

+) \(x^2+x+6=0\)

Ta có: \(\Delta=1^2-4.6=-25< 0\)

Vậy pt có 2 nghiệm\(\left\{1;-2\right\}\)

a, \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

\(\Leftrightarrow x^4+2x^3+x^2+4x^2+4x+12=0\)

\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^3+2x^2+x^2+2x+6x+12\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)

có : \(x^2+x+6>0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)

b,  \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)

\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]-297=0\)

\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+7x-21\right)-297=0\)

đặt \(x^2+4x-13=t\)

\(\Leftrightarrow\left(t+8\right)\left(t-8\right)-297=0\)

\(\Leftrightarrow t^2-64-297=0\)

\(\Leftrightarrow t^2=361\)

\(\Leftrightarrow t=\pm19\)

có t rồi tìm x thôi

Ta có : \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

            \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

  \(\Leftrightarrow\left(x^2+x\right)+2.2.\left(x^2+x\right)+4-16=0\)

\(\Leftrightarrow\left[\left(x^2+x\right)^2+2.2\left(x^2+x\right)+4\right]=16\)

\(\Leftrightarrow\left(x^2+x+2\right)^2=16\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+x+2=4\\x^2+x+2=-4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1;-2\\vônghiệm\end{array}\right.\)

Vậy \(S=\left\{-2;1\right\}\)

Ta có : \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

Đặt \(t=x^2+x\) , pt trở thành \(t^2+4t-12=0\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=2\\t=-6\end{array}\right.\)

Nếu t = 2 ta có pt : \(x^2+x=2\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

Nếu t = -6 , ta có pt : \(x^2+x=-6\Leftrightarrow x^2+x+6=0\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)

mà \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}>0\) . Dấu đẳng thức không xảy ra nên pt này vô nghiệm.

Vậy tập nghiệm của pt : S={-2;1}

giải phương trình  : (x²+x)²+4(x²+x)=12

<=>\(\left(x^2+x+2\right)^2-12-4=0\)

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

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

<=> \(x^2+x-6=0,x^2+x+2=0\)

pt1: \(x^2+x-6=0\)

<=>x=2 V x=-3

PT2: x^2+x+2=0    vô nghiệm

=> taaoj nghiệm pt S={-3,2}

tôi chịu

b)  Đặt  \(x-7=a\) ta có:

         \(\left(a+1\right)^4+\left(a-1\right)^4=16\)

 \(\Leftrightarrow\)\(a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16\)

 \(\Leftrightarrow\)\(2a^4+12a^2+2-16=0\)

 \(\Leftrightarrow\)\(2\left(a^4+6a^2-7\right)=0\)

 \(\Leftrightarrow\)\(a^4+6a^2-7=0\)

 \(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)

Vì     \(a^2+7>0\) nên    \(\orbr{\begin{cases}a-1=0\\a+1=0\end{cases}}\)

Thay trở lại ta có:   \(\orbr{\begin{cases}x-8=0\\x-6=0\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)

Vậy...

b) \(\left(x-6\right)^4+\left(x-8\right)^4=16\)

Ta có: \(\left(x-6\right)^4+\left(x-8\right)^4=16\)(1)

Đặt t = x - 7, từ (1) suy ra:

\(\Leftrightarrow\left(t^4+4t^3+6t^2+4t+1\right)+\left(t^3-4t^3+6t^2-4t+1\right)\)

\(\Leftrightarrow2t^4+12t^2+2=16\)

\(\Leftrightarrow t^4+6t^2+1=8\)

\(\Leftrightarrow t^4+6t^2-7=0\)

\(\Leftrightarrow\left(t^4-1\right)+\left(6t^2-6\right)=0\)

\(\Leftrightarrow\left(t^2+1\right)\left(t^2-1\right)+6.\left(t^2-1\right)=0\)

\(\Leftrightarrow\left(t^2-1\right)\left(t^2+1+6\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)\left(t^2+7\right)=0\)

Vì: \(t^2+7\ge7\)nên:

\(\left(t-1\right)\left(t+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=8\end{cases}}\)

\(\Rightarrow x\in\left\{6;8\right\}\)

Đặt \(t=x^2+x\) ta có pt sau: 

\(t^2+4t=12\Rightarrow t^2+4t-12=0\)

\(\Rightarrow t^2-2t+6t-12=0\)

\(\Rightarrow t\left(t-2\right)+6\left(t-2\right)=0\)

\(\Rightarrow\left(t-2\right)\left(t+6\right)=0\)\(\Rightarrow\orbr{\begin{cases}t=2\\t=-6\end{cases}}\)

*)Xét \(x^2+x=2\Rightarrow x^2+x-2=0\)

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

*)Xét \(x^2+x=-6\Rightarrow x^2+x+6=0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\) (vô nghiệm)