Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\) ta được:
a/ \(x^2+\frac{1}{x^2}+6\left(x+\frac{1}{x}\right)+11=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(\Leftrightarrow t^2-2+6t+11=0\Leftrightarrow\left(t+3\right)^2=0\)
\(\Rightarrow t=-3\Rightarrow x+\frac{1}{x}=-3\Leftrightarrow x^2+3x+1=0\) (casio)
b/ \(x^2+\frac{1}{x^2}-10\left(x+\frac{1}{x}\right)+26=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(\Leftrightarrow t^2-2-10t+26=0\)
\(\Leftrightarrow t^2-10t+24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{x}=4\\x+\frac{1}{x}=6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x=1=0\\x^2-6x+1=0\end{matrix}\right.\) (casio)
2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)
\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)
\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-5x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)
1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)
\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+3\right)+\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x^2+x+3\right)=0\)
mà \(x^2+x+3>0\forall x\)
nên (x+1)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy: S={-1;-3}
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 \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
a/ Nhận thấy \(x=0\) không phải nghiệm, chia cả 2 vế của pt cho \(x^2\):
\(x^2+5x-10+\frac{10}{x}+\frac{4}{x^2}=0\)
\(\Leftrightarrow x^2+\frac{4}{x^2}+5\left(x+\frac{2}{x}\right)-10=0\)
Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+4+\frac{4}{x^2}=a^2\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)
Phương trình trở thành:
\(a^2-4+5a-10=0\)
\(\Leftrightarrow a^2+5a-14=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=2\\x+\frac{2}{x}=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+2=0\left(vn\right)\\x^2+7x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-7+\sqrt{41}}{2}\\x=\frac{-7-\sqrt{41}}{2}\end{matrix}\right.\)
b/ \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow x^4-8x^2+16+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(x^2-4=a\Rightarrow-4=a-x^2\)
Phương trình trở thành:
\(a^2+x+a-x^2=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x\right)+x+a=0\)
\(\Leftrightarrow\left(a-x+1\right)\left(x+a\right)=0\)
\(\Leftrightarrow\left(x^2-4-x+1\right)\left(x+x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\x^2+x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{13}}{2}\\x=\frac{-1\pm\sqrt{17}}{2}\end{matrix}\right.\)
1. phương trình tương đương với \(\left(x^2-7x+2\right)\left(x^2+2x+2\right)=0\to x=\frac{7}{2}\pm\frac{\sqrt{41}}{2}\)
2. phương trình tương đương với \(\left(x^2+\left(\sqrt{2}-1\right)x+1\right)\left(x^2+\left(\sqrt{2}+1\right)x-1\right)=0\to x=\frac{-1\pm\sqrt{2}\pm\sqrt{7-2\sqrt{2}}}{2}\) với dấu +,- lấy tuỳ ý
a)\(x^4-8x^2+x+12=0\)
\(\Leftrightarrow x^4-x^3-3x^2+x^3-x^2-3x-4x^2+4x+12=0\)
\(\Leftrightarrow x^2\left(x^2-x-3\right)+x\left(x^2-x-3\right)-4\left(x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3\right)\left(x^2+x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-x-3=0\\x^2+x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\Delta\left(1\right)=\left(-1\right)^2-\left(-4\left(1\cdot3\right)\right)=13\\\Delta\left(2\right)=1^2-\left(-4\left(1\cdot4\right)\right)=17\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x_{1,2}=\frac{1\pm\sqrt{13}}{2}\\x_{1,2}=\frac{-1\pm\sqrt{17}}{2}\end{cases}}\)
b)\(x^4+5x^3-10x^2+10x+4=0\)
\(\Leftrightarrow x^4-2x^3+2x^2+7x^3-14x^2+14x+2x^2-4x+4=0\)
\(\Leftrightarrow x^2\left(x^2-2x+2\right)+7x\left(x^2-2x+2\right)+2\left(x^2-2x+2\right)=0\)
\(\Leftrightarrow\left(x^2-2x+2\right)\left(x^2+7x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-2x+2=0\\x^2+7x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\Delta\left(1\right)=\left(-2\right)^2-4\cdot1\cdot2=-4< 0\left(loai\right)\\\Delta\left(2\right)=7^2-4\cdot1\cdot2=41\end{cases}}\)\(\Rightarrow x_{1,2}=\frac{-7\pm\sqrt{41}}{2}\)
bạn dùng hệ số bất định
(x2+ax+b)(x2+cx+d)=x4+cx3+dx2+ax3+acx2+adx+bx2+bcx+bd
=x4+x3(a+c)+x2(b+ac+d)+x(ad+bc)+bd
=>a+c=-1
=>b+ac+d=-10 =>a=2;b=-2;c=-3;d=-2
=>ad+bc=20
=>bd=4
vây x4-x3-10x2+20x+4=(x2+2x-2)(x2-3x-2)=0
=> x2+2x-2=0
=> x2-3x-2=0 bạn tự giải nhé
\(\left(x^2+\text{ax}+b\right)\left(x^2+cx+d\right)=x^4+cx^3+dx^2+\text{ax}^3+acx^2+adx+bx^2+bcx+bd\\ =>a+c=1\\ =>b+ac+d=-10\)
\(=>ad+bc=20\\ =>a=2;b=-2;c=-3;d=-2\\ =>bd=4\\ \)
Vậy \(x^4-x^3-10x^2+20x+4=\left(x^2+2x-2\right)\left(x^2-3x-2\right)=0\\ =>x^2+2x-2=0\\ =>x^2-3x-2=0\)
\(=>x^2-x-2x-2=0\\ =>x\left(x-1\right)-2\left(x-1\right)=0\\ =>\left(x-1\right)\left(x-2\right)=0\)
tới đây chắc dễ dàng
`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`
2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`