Giải pt
x^4-4x^2+8x+4=0
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HP
1
AT
24 tháng 2 2018
ĐKXĐ:\(x\ne\pm\dfrac{1}{2}\)
\(\dfrac{1+8x}{4+8x}-\dfrac{4x}{12x-6}+\dfrac{32x^2}{3\left(4-16x^2\right)}=0\)
\(\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{4x}{6\left(2x-1\right)}+\dfrac{32x^2}{-6\cdot\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\dfrac{6\cdot\left(1+8x\right)\left(2x-1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{4\cdot4x\left(2x+1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{32x^2\cdot4}{24\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow96x^2-36x-6-36x^2-16x-144x^2=0\)
\(\Leftrightarrow-84x^2-52x-6=0\)
\(\Leftrightarrow\Delta=688\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{52-\sqrt{688}}{-168}=\dfrac{-13+\sqrt{43}}{42}\\x_2=\dfrac{52+\sqrt{688}}{-168}=\dfrac{-13-\sqrt{43}}{43}\end{matrix}\right.\)
Vậy pt có 2 nghiệm phân biệt............
28 tháng 8 2021
a: Ta có: \(x^2+3x+4=0\)
\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)
Do đó: Phương trình vô nghiệm
16 tháng 12 2021
anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2
TL
2
24 tháng 1 2021
(4x - 3)2 - (2x + 1)2 = 0
\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0
\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
3x - 12 - 5x(x - 4) = 0
\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0
\(\Leftrightarrow\) -5x2 + 23x - 12 = 0
\(\Leftrightarrow\) 5x2 - 23x + 12 = 0
\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0
\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0
\(\Leftrightarrow\) (x - 4)(5x - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
(8x + 2)(x2 + 5)(x2 - 4) = 0
\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0
Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x
\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
24 tháng 1 2021
a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)
b) Ta có: \(3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)
c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)
mà \(2>0\)
và \(x^2+5>0\forall x\)
nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 6 2019
\(x^4-4x^3-5x^2-3x^2+12x+15=0\)
\(\Leftrightarrow x^2\left(x^2-4x-5\right)-3\left(x^2-4x-5\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2-4x-5\right)=0\)
21 tháng 6 2019
\(x^4-4x^3-8x^2+12x+15=0\)
\(\Leftrightarrow x^4+x^3-5x^3-5x^2-3x^2-3x+15x+15=0\)
\(\Leftrightarrow x^3\left(x+1\right)-5x^2\left(x+1\right)-3x\left(x+1\right)+15\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-5x^2-3x+15\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x-5\right)-3\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\\x=\pm\sqrt{3}\end{matrix}\right.\)
TD
0
28 tháng 9 2020
Ta có: \(8x^4-8x^3-4x^2+3x+1=0\)
\(\Leftrightarrow8x^3\left(x-1\right)-\left(4x^2-3x-1\right)=0\)
\(\Leftrightarrow8x^3\left(x-1\right)-\left(4x^2-4x+x-1\right)=0\)
\(\Leftrightarrow8x^3\left(x-1\right)-\left[4x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow8x^3\left(x-1\right)-\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(8x^3-4x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(8x^3+4x^2-4x^2-2x-2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[4x^2\left(2x+1\right)-2x\left(2x+1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)\left(4x^2-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\\4x^2-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-1\\\left(2x\right)^2-2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}-\frac{5}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\\\left(2x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\\2x-\frac{1}{2}=\frac{\sqrt{5}}{2}\\2x-\frac{1}{2}=\frac{-\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\\2x=\frac{\sqrt{5}+1}{2}\\2x=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\\x=\frac{\sqrt{5}+1}{4}\\x=\frac{1-\sqrt{5}}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\frac{1}{2};\frac{\sqrt{5}+1}{4};\frac{1-\sqrt{5}}{4}\right\}\)
3 tháng 2 2019
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)
\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^2+3x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)
.......................................................................................
\(x^3-8x^2-8x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)
......................................................................................
\(x^4-4x^2+8x+4=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+8\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x+2\right)+8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-2x^2+8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x^3-2x^2+8=0\end{array}\right.\)
Tới đây tự giải nhé :)
Đầu tiên ta phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
Suy ra pt : \(\left(x^2-2x+2\right)\left(x^2+2x+2\right)-4x\left(x-2\right)=0\)
Nhận thấy x = 0 không là nghiệm của pt, do đó chia cả hai vế của pt cho \(x^4\ne0\) được :
\(\left(1-\frac{2}{x}+\frac{2}{x^2}\right)\left(1+\frac{2}{x}+\frac{2}{x^2}\right)-4\left(\frac{1}{x^2}-\frac{2}{x^3}\right)=0\)
Đặt \(t=\frac{2}{x}\) , pt trở thành : \(\left(1-2t+2t^2\right)\left(1+2t+2t^2\right)-4\left(t^2-2t^3\right)=0\)
Tới đây thử giải pt với ẩn t xem có đc k