\(\Rightarrow\)[ (x-1)(x+5) ].[ (x-3)(x+7) ] = 297

\(\Rightarrow\)( x^2 + 4x - 5 ) . ( x^2 + 4x - 21) = 297 ( bước này mình làm tắt)

\(\Rightarrow\)(x^2 + 4x - 13 + 8) . (x^2 + 4x -13 - 8) = 297

\(\Rightarrow\)(x^2 + 4x - 13)^2 - 64 = 297

\(\Rightarrow\)(x^2 + 4x -13)^2 = 361

\(\Rightarrow\)x^2 + 4x - 13 = 19 hoặc x^ + 4x -13 = -19

phần còn lại tự bạn giải nha

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

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

Đặt    \(x^2+4x-5=t\)   ta có:

                    \(t\left(t-16\right)-297=0\)

       \(\Leftrightarrow\)\(t^2-16t+64-361=0\)

       \(\Leftrightarrow\) \(\left(t-8\right)^2-361=0\)

       \(\Leftrightarrow\)\(\left(t-8-19\right)\left(t-8+19\right)=0\)

       \(\Leftrightarrow\)\(\left(t-27\right)\left(t+11\right)=0\)

       \(\Leftrightarrow\)\(\orbr{\begin{cases}t-27=0\\t+11=0\end{cases}}\)

Thay trở lại ta được:    \(\orbr{\begin{cases}x^2+4x-32=0\\x^2+4x+6=0\end{cases}}\)

                        \(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-4\right)\left(x+8\right)=0\\\left(x+2\right)^2+2=0\left(L\right)\end{cases}}\)

                       \(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)

Vậy...

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

2) \(\left(x-2\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)

3) \(\left(7-x\right)\left(x+19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)

4) \(-5< x< 1\)

\(\Rightarrow x\in\left\{-1;-3;-2;-1;0\right\}\)

5) \(\left(x-3\right)\left(x-5\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}x-3>0\\x-5< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>3\\x< 5\end{matrix}\right.\)

\(\Rightarrow3< x< 5\)

6) \(2x^2-3=29\)

\(\Rightarrow2x^2=29+3\)

\(\Rightarrow2x^2=32\)

\(\Rightarrow x^2=\dfrac{32}{2}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

7) \(-6x-\left(-7\right)=25\)

\(\Rightarrow-6x+7=25\)

\(\Rightarrow-6x=25-7\)

\(\Rightarrow-6x=18\)

\(\Rightarrow x=\dfrac{18}{-6}\)

\(\Rightarrow x=-3\)

8) \(46-\left(x-11\right)=-48\)

\(\Rightarrow x-11=48+46\)

\(\Rightarrow x-11=94\)

\(\Rightarrow x=94+11\)

\(\Rightarrow x=105\)

1: (x-2)(x+4)=0

=>x-2=0 hoặc x+4=0

=>x=2 hoặc x=-4

2: (x-2)(x+15)=0

=>x-2=0 hoặc x+15=0

=>x=2 hoặc x=-15

3: (7-x)(x+19)=0

=>7-x=0 hoặc x+19=0

=>x=7 hoặc x=-19

4: -5<x<1

=>\(x\in\left\{-4;-3;-2;-1;0\right\}\)

5: (x-3)(x-5)<0

=>x-3>0 và x-5<0

=>3<x<5

6: 2x^2-3=29

=>2x^2=32

=>x^2=16

=>x=4 hoặc x=-4

7: -6x-(-7)=25

=>-6x=25-7=18

=>x=-3

8: 46-(x-11)=-48

=>x-11=46+48=94

=>x=94+11=105

Ta có:\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)

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

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

Đặt \(x^2+4x-5=t\) thì \(t\left(t-16\right)=297\)

\(\Leftrightarrow t^2-16t-297=0\Leftrightarrow t^2-27t+11t-297=0\)

\(\Leftrightarrow t\left(t-27\right)+11\left(t-27\right)=0\Leftrightarrow\left(t+11\right)\left(t-27\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=-11\\t=27\end{cases}}\)

Với \(t=-11\) thì \(x^2+4x-5=-11\Leftrightarrow x^2+4x+6=0\Leftrightarrow x^2+4x+4+2=0\)

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

Với \(t=27\) thì \(x^2+4x-5=27\Leftrightarrow x^2+4x-32=0\Leftrightarrow x^2-4x+8x-32=0\)

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

Tập nghiệm của pt \(S=\left\{-8,4\right\}\)

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

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

\(\Leftrightarrow\left(x^2+5x-x-5\right)\left(x^2+7x-3x-21\right)=297\)

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

Đặt \(x^2+4x-13=m\)

Ta có : \(\left(m+8\right)\left(m-8\right)=297\)

\(\Leftrightarrow m^2-8^2=297\)

\(\Leftrightarrow m^2=361\)

\(\Leftrightarrow m=\pm19\)

+) Với m = 19 ta có : \(x^2+4x-13=19\)

\(\Leftrightarrow x^2+4x-32=0\)

\(\Leftrightarrow\left(x^2-4x\right)+\left(8x-32\right)=0\)

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

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

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

+) Với m = -19 ta có : \(x^2+4x-13=-19\)

\(\Leftrightarrow x^2+4x+6=0\)

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

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

\(\Leftrightarrow\left(x+2\right)^2=-2\) ( vô lí )

Vậy phương trình có tập nghiệm \(S=\left\{4;-8\right\}\)

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

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

Đặt \(t=x^2+4x-5\) ta có:

\(\left(t-16\right)t-297=0\)\(\Rightarrow t^2-16t-297=0\)

\(\Rightarrow t^2-27t+11t-297=0\)

\(\Rightarrow t\left(t-27\right)+11\left(t-27\right)=0\)

\(\Rightarrow\left(t+11\right)\left(t-27\right)=0\)\(\Rightarrow\left[\begin{matrix}t=-11\\t=27\end{matrix}\right.\)

*)Xét \(t=-11\Rightarrow x^2+4x-5=-11\)

\(\Rightarrow x^2+4x+6=0\Rightarrow\left(x+2\right)^2+2>0\left(loai\right)\)

*)Xét \(t=27\Rightarrow x^2+4x-5=27\)

\(\Rightarrow x^2+4x-32=0\Rightarrow\left(x+8\right)\left(x-4\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-8\\x=4\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(x^2+5x-x-5\right)\left(x^2+7x-3x-21\right)=297\)

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

Đặt \(x^2+4x-13=y\)

Ta có phương trình (*) \(\Leftrightarrow\left(y+8\right)\left(y-8\right)=297\)

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

\(\Leftrightarrow y^2-361=0\Leftrightarrow\left(y-19\right)\left(y+19\right)=0\)

\(\Leftrightarrow\left(x^2+4x-13-19\right)\left(x^2+4x-13+19\right)=0\)

\(\Leftrightarrow\left(x^2+4x-32\right)\left(x^2+4x+6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+8\right)\left(x^2+4x+6\right)=0\)

Ta có: \(x^2+4x+6=\left(x^2+4x+4\right)+2=\left(x+2\right)^2+2\)

Vì \(\left(x+2\right)^2\ge0\forall x\Rightarrow\left(x+2\right)^2+2\ge2>0\forall x\)

\(\Rightarrow\) Phương trình \(x^2+4x+6\) vô nghiệm.

Vậy \(\left(x-4\right)\left(x+8\right)\left(x^2+4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

Vậy phương trình trên có tập nghiệm là S = {4:-8}

(x-1)(x-3)(x+5)(x+7)=297

⇔(x-1)((x+5)(x-3)(x+7)=297

⇔(x²+4x-5)(x²-4x-21)=297

Đặt x²+4x-13=t, ta được:

(t+8)(t-8)=297

⇔t²-64=297

⇔t²-64-297=0

⇔t²-361=0

⇔(t-19)(t+19)=0

\(\left\{{}\begin{matrix}t-19=0\\t+19=0\end{matrix}\right.< =>\left\{{}\begin{matrix}t=19\\t=-19\end{matrix}\right.\)

Với t=19, ta được:

x²+4x-13=19

⇔x²+4x-13-19=0

⇔x²+4x-32=0

⇔x²+8x-4x-32=0

⇔x(x+8)-4(x+8)=0

⇔(x+8)(x-4)=0

⇔\(\left\{{}\begin{matrix}x+8=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\x=4\end{matrix}\right.\)

->phương trình có tập nghiệm là S=\(\left\{-8;4\right\}\)

Với t=-19,ta được:

x²+4x-13=-19

⇔x²+4x-13+19=0

⇔x²+4x+6=0

⇔x²+4x+4+2=0

⇔(x+2)²+2=0 (vì (x+2)²≥0 với ∀ ⇒(x+2)²+2 ≥ 2 >0)

->Phương trình vô nghiệm

Kết luận : Vậy phương trình có tập nghệm là S=\(\left\{-8;4\right\}\)

d: \(x\left(x+1\right)\left(x^2+x+1\right)=42\left(1\right)\)

=>\(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(a=x^2+x\)

Phương trình (1) sẽ trở thành \(a\left(a+1\right)=42\)

=>\(a^2+a-42=0\)

=>(a+7)(a-6)=0

=>\(\left(x^2+x+7\right)\left(x^2+x-6\right)=0\)

mà \(x^2+x+7=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}>0\forall x\)

nên \(x^2+x-6=0\)

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

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

e: \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\left(2\right)\)

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

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

Đặt \(b=x^2+4x\)

Phương trình (2) sẽ trở thành \(\left(b-5\right)\left(b-21\right)-297=0\)

=>\(b^2-26b+105-297=0\)

=>\(b^2-26b-192=0\)

=>(b-32)(b+6)=0

=>\(\left(x^2+4x-32\right)\left(x^2+4x+6\right)=0\)

mà \(x^2+4x+6=\left(x+2\right)^2+2>0\forall x\)

nên \(x^2+4x-32=0\)

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

=>\(\left[{}\begin{matrix}x+8=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=4\end{matrix}\right.\)

f: \(x^4-2x^2-144x-1295=0\)

=>\(x^4-7x^3+7x^3-49x^2+47x^2-329x+185x-1295=0\)

=>\(\left(x-7\right)\cdot\left(x^3+7x^2+47x+185\right)=0\)

=>\(\left(x-7\right)\left(x+5\right)\left(x^2+2x+37\right)=0\)

mà \(x^2+2x+37=\left(x+1\right)^2+36>0\forall x\)

nên (x-7)(x+5)=0

=>\(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

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

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

\(\Rightarrow\left[x^2+4x-5\right]\left[x^2+4x-5-16\right]=297\)

Đặt \(x^2+4x-5=t\)

\(\Rightarrow t\left(t-16\right)=297\)

\(\Rightarrow t^2-16t+64=297+64\)

\(\Rightarrow\left(t+8\right)^2=361\)

\(\Rightarrow\left[{}\begin{matrix}t+8=19\\t+8=-19\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=11\\t=-27\end{matrix}\right.\)

Ta có : \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)

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

\(\Leftrightarrow\left(x^2-x+5x-5\right)\left(x^2-3x+7x-21\right)=297\)

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

\(\Leftrightarrow\left(x^2+4x-13+8\right)\left(x^2+4x-13-8\right)=297\)

\(\Leftrightarrow\left(x^2+4x-13\right)^2-64=297\)

\(\Leftrightarrow\left(x^2+4x-13\right)^2=361\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x-13=19\\x^2+4x-13=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+4-17=19\\x^2+4x+4-17=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2-17=19\\\left(x+2\right)^2-17=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2=36\\\left(x+2\right)^2=-2\left(VL\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)