\(\)x4 - 2x2 + 144x- 1295=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
=>\(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-2x^2-144x-1295=0\)
\(\Leftrightarrow x^4-7x^3+7x^3-49x^2+47x^2-329x+185x-1295=0\)
\(\Leftrightarrow x^3\left(x-7\right)+7x^2\left(x-7\right)+47x\left(x-7\right)+185\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x^3+7x^2+47x+185\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x^3+5x^2+2x^2+10x+37x+185\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left[x^2\left(x+5\right)+2x\left(x+5\right)+37\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+5\right)\left(x^2+2x+37\right)=0\)
\(\Leftrightarrow\)\(x-7=0\)
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 \(S=\left\{7;-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-2x^2-144x-1295=0\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(4x^2+144x+1296\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-\left(2x+36\right)^2=0\)
\(\Leftrightarrow\left(x^2+1+2x+36\right)\left[x^2+1-\left(2x+36\right)\right]=0\)
\(\Leftrightarrow\left(x^2+2x+37\right)\left(x^2-2x-35\right)=0\)
\(\Leftrightarrow\left(x^2+5x-7x-35\right)\left(x^2+2x+1+36\right)=0\)
\(\Leftrightarrow\left[x\left(x+5\right)-7\left(x+5\right)\right]\left[\left(x+1\right)^2+36\right]=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-7\right)\left[\left(x+1\right)^2+36\right]=0\)
Dễ thấy:\(\left(x+1\right)^2+36\ge36>0\forall x\) (vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=7\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x4−2x2−144x−1295=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+37)(x2−2x−35)=0
⇔(x2+5x−7x−35)(x2+2x+1+36)=0
⇔[x(x+5)−7(x+5)][(x+1)2+36]=0
⇔(x+5)(x−7)[(x+1)2+36]=0
Dễ thấy:(x+1)2+36≥36>0∀x (vô nghiệm)
⇒[x+5=0x−7=0
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 \(S=\left\{-5;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
pt <=> (x^4-5x^3)+(5x^3-25x^2)+(23x^2-115x)+(259x-1295) = 0
<=> (x-5).(x^3+5x^2+23x+259) = 0
<=> (x-5).[(x^3+7x^2)-(2x^2+14x)+(37x+259)] = 0
<=> (x-5).(x+7).(x^2-2x+37) = 0
<=> (x-5).(x+7) = 0 ( vì x^2-2x+37 > 0 )
<=> x-5=0 hoặc x+7=0
<=> x=5 hoặc x=-7
Vậy .........
Tk mk nha