=>x^4+2x^2+1-4x^2-144x-1296=0

=>(x^2+1)^2-(2x+36)^2=0

=>(x^2+1-2x-36)(x^2+1+2x+36)=0

=>x^2-2x-35=0

=>(x-7)(x+5)=0

=>x=7 hoặc x=-5

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

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

=>x^4+2x^2+1-4x^2-144x-1296=0

=>(x^2+1)^2-(2x+36)^2=0

=>(x^2+1-2x-36)(x^2+1+2x+36)=0

=>x^2-2x-35=0

=>(x-7)(x+5)=0

=>x=7 hoặc x=-5

\(x^4-2x^2-144x-1295=0\)

\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(4x^2+144x+1296\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)^2-\left(2x+36\right)^2=0\)

\(\Leftrightarrow\left(x^2+1+2x+36\right)\left[x^2+1-\left(2x+36\right)\right]=0\)

\(\Leftrightarrow\left(x^2+2x+37\right)\left(x^2-2x-35\right)=0\)

\(\Leftrightarrow\left(x^2+5x-7x-35\right)\left(x^2+2x+1+36\right)=0\)

\(\Leftrightarrow\left[x\left(x+5\right)-7\left(x+5\right)\right]\left[\left(x+1\right)^2+36\right]=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-7\right)\left[\left(x+1\right)^2+36\right]=0\)

Dễ thấy:\(\left(x+1\right)^2+36\ge36>0\forall x\) (vô nghiệm)

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

Bài làm

x⁴ - 2x² - 144x - 1295 = 0

<=> x⁴ + 2x² - 4x² - 144x - 1296 + 1 = 0

<=> ( x⁴ + 2x² + 1 ) - ( 4x² + 144x + 1296 ) = 0

<=> ( x² + 1 )² - ( 2x + 36 )² = 0

<=> ( x² + 1 - 2x - 36 )( x² + 1 + 2x + 36 ) = 0

<=> ( x² - 2x - 35 )( x² + 2x + 37 ) = 0

<=> ( x² + 5x - 7x - 35 ) ( x² + 2x + 1 + 36 ) = 0

<=> [ x( x + 5 ) - 7( x + 5 ) ][ ( x² + 2x + 1 ) + 36 ] = 0

<=> ( x + 5 )( x - 7 )[ ( x + 1 )² + 36 ] = 0

Ta có: ( x + 1 )² \(\ge\)0

<=> ( x + 1 )² + 36 \(\ge\)36

=> ( x + 5 )( x - 7 ) = 0

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

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

Vậy tập nghiệm phương trình trên là: S = { -5; 7 }

# Học tốt #

Bài này không nhất thiết phải đặt ẩn phụ nên mình giải luôn nhé.

\(x^4-2x^2-144x-1295=0\\ \Leftrightarrow x^4-4x^2+2x^2-144x+1-1296=0\\\Leftrightarrow \left(x^4+2x^2+1\right)-\left(4x^2+144x+1296\right)=0\\ \Leftrightarrow\left(x^2+1\right)^2-\left(2x+36\right)=0\\ \Leftrightarrow\left(x^2+1-2x-36\right)\left(x^2+1+2x+36\right)=0\\ \Leftrightarrow\left(x^2-2x-35\right)\left[\left(x+1\right)^2+36\right]=0\\\Leftrightarrow \left(x^2+5x-7x-35\right)\left[\left(x+1\right)^2+36\right]=0\\ \Rightarrow\left(x+5\right)\left(x-7\right)\left[\left(x+1\right)^2+36\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=7\end{matrix}\right.\left(vi\left[\left(x+1\right)^2+36>0\forall\right]x\right)\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-5;7\right\}\)

$x^4-2x^2-144x-1295=0$

$\iff \left(x^4+2x^2+1\right)-\left(4x^2+144x+1296\right)=0$

$\iff {\left(x^2+1\right)}^2-{\left(2x+36\right)}^2=0$

$\iff \left(x^2+1+2x+36\right)\left[x^2+1-\left(2x+36\right)\right]=0$

$\iff \left(x^2+2x+37\right)\left(x^2-2x-35\right)=0$

$\iff \left(x^2+5x-7x-35\right)\left(x^2+2x+1+36\right)=0$

$\iff \left[x\left(x+5\right)-7\left(x+5\right)\right]\left[{\left(x+1\right)}^2+36\right]=0$

$\iff \left(x+5\right)\left(x-7\right)\left[{\left(x+1\right)}^2+36\right]=0$

Dễ thấy:${\left(x+1\right)}^2+36\ge 36>0\forall x$ (vô nghiệm)

$\Rightarrow [\begin{array}{c}x+5=0\\ x-7=0\end{array}$$\Rightarrow [\begin{array}{c}x=-5\\ x=7\end{array}$

\(x^4-2x^2-144x-1295=0\)

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

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

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

\(\Leftrightarrow\left(x-7\right)\left(x^3+5x^2+2x^2+10x+37x+185\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left[x^2\left(x+5\right)+2x\left(x+5\right)+37\left(x+5\right)\right]=0\)

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

\(\Leftrightarrow\)\(x-7=0\)

hoặc   \(x+5=0\)

hoặc   \(x^2+2x+37=0\)

\(\Leftrightarrow\)\(x=7\)(tm)

hoặc   \(x=-5\)(tm)

hoặc \(\left(x+1\right)^2+36=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{7;-5\right\}\)

Bạn vào link này:

Câu hỏi của Nguyễn Thị Thu Trng

Chúc bạn học tốt!!!

1) \(x^4-2x^2-144x+1295=0\)

\(\Rightarrow\)Cậu xem lại đề thử xem nhé !

2) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-1\right)-24=0\)

\(\Leftrightarrow x^4+2x^3-x^2-2x-24=0\)

\(\Leftrightarrow x^4+x^3+4x^2+x^3+x^2+4x-6x^2-6x-24=0\)

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

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

\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x^2+x+4\right)=0\)

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

\(\Leftrightarrow\)\(x+3=0\)

hoặc \(x-2=0\)

hoặc \(x^2+x+4=0\)

\(\Leftrightarrow\)\(x=-3\left(tm\right)\)

hoặc   \(x=2\left(tm\right)\)

hoặc  \(\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

3) \(x^4-2x^3+4x^2-3x-10=0\)

\(\Leftrightarrow x^4+x^3-3x^3-3x^2+7x^2+7x-10x-10=0\)

\(\Leftrightarrow x^3\left(x+1\right)-3x^2\left(x+1\right)+7x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-3x^2+7x-10\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-2x^2-x^2+2x+5x-10\right)=0\)

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

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

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

hoặc \(x-2=0\)

hoặc \(x^2-x+5=0\)

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

hoặc \(x=2\left(tm\right)\)

hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-1;2\right\}\)