K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
1 tháng 7 2019
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow x^2+7x+1-x^3-7x^2-x=0\)
\(\Leftrightarrow\left(x^2+7x+1\right)-x\left(x^2+7x+1\right)=0\)
\(\Leftrightarrow\left(x^2+7x+1\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left[\left(x^2+7x+\frac{49}{4}\right)-\frac{45}{4}\right]\left(1-x\right)=0\)
\(\Leftrightarrow\left[\left(x+\frac{7}{2}\right)^2-\frac{45}{4}\right]\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{45}-7}{2}\\x=-\frac{\sqrt{45}+7}{2}\\x=1\end{matrix}\right.\)
NT
14 tháng 8 2018
2x4 - 6x3 + x2 + 6x - 3 = 0
=> 2x4 - 2x3 - 4x3 + 4x2 - 3x2 + 3x + 3x - 3 = 0
=> 2x3(x - 1) - 4x2(x - 1) - 3x(x - 1) + 3(x - 1) = 0
=> (x - 1)(2x3 - 4x2 - 3x + 3) = 0
=> (x - 1)(2x3 + 2x2 - 6x2 - 6x + 3x + 3) = 0
=> (x - 1)[2x2(x + 1) - 6x(x + 1) + 3(x + 1)] = 0
=> (x - 1)(x + 1)(2x2 - 6x + 3) = 0
\(\Rightarrow\left[{}\begin{matrix}x-1\\x+1\\2x^2-6x+3\end{matrix}\right.\) (2x2 - 6x + 3 vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
TN
23 tháng 7 2017
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
2 tháng 3 2018
pt <=> 3x^2-6x+4y^2 = 13
<=> (3x^2-6x+3)+4y^2 = 16
<=> 3.(x-1)^2+4y^2 = 16
<=> 3.(x-1)^2 < = 16
<=> (x-1)^2 < = 16/3
Mà (x-1)^2 > = 0
=> 0 < = (x-1)^2 < = 16/3
Mặt khác x thuộc Z nên x-1 thuộc Z => (x-1)^2 thuộc N
=> (x-1)^2 thuộc {0;1;4}
Đến đó bạn tự tìm x,y nha
Tk mk nha
TA
4
DT
0
25 tháng 7 2019
1. \(\left(5x-1\right)\left(x-1\right)+6x-44=0\)
=> \(5x^2-5x-x+1+6x-44=0\)
=> \(5x^2-43=0\)
=> \(5x^2=43\)
=> \(x^2=43:5\)
=> \(x^2=\frac{43}{5}\)
=> \(\orbr{\begin{cases}x=\sqrt{\frac{43}{5}}\\x=-\sqrt{\frac{43}{5}}\end{cases}}\)
2. \(\left(2x+1\right)\left(4x^2-2x+1\right)+9=0\)
=> \(8x^3+1+9=0\)
=> \(8x^3+10=0\)
=> \(8x^3=-10\)
=> \(x^3=-10:8\)
=> \(x^3=-\frac{5}{4}\)
=> ko bt
25 tháng 7 2019
2,\(\left(2x+1\right)\left(4x^2-2x+1\right)+9=0\)
\(\Rightarrow8x^3-4x^2+4x^2-2x+1+9=0\)
\(\Rightarrow8x^3+10=0\)
\(\Rightarrow x^3=\frac{-5}{4}\)
\(\Rightarrow x=\sqrt[3]{\frac{-5}{4}}\)
Vậy....
CN
22 tháng 12 2017
a) \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b) (2x - 1)2 - (x + 3)2 = 0
⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0
⇔ (x - 4)(3x +2) = 0
⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)
c) 2x2 - x - 6 = 0
⇔ 2x2 - 4x + 3x - 6 = 0
⇔ 2x(x - 2) + 3(x - 2) = 0
⇔ (x - 2) (2x + 3) = 0
⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)
NT
22 tháng 12 2017
2)a.
\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)
b)
\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow-x^3-6x^2+6x+1=0\)
\(\Leftrightarrow-x^3+x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow-x^2\left(x-1\right)-7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2-7x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-x^2-7x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+12,25-11,25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5\right)^2-\left(\frac{3\sqrt{5}}{2}\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5-\frac{3\sqrt{5}}{2}\right)\left(x+3,5+\frac{3\sqrt{5}}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+3,5-\frac{3\sqrt{5}}{2}=0\\x+3,5+\frac{3\sqrt{5}}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3,5+\frac{3\sqrt{5}}{2}=\frac{-7+3\sqrt{5}}{2}\\x=-3,5-\frac{3\sqrt{5}}{2}=\frac{-7-3\sqrt{5}}{2}\end{matrix}\right.\)
Vậy x = \(\left\{1;\frac{-7+3\sqrt{5}}{2};\frac{-7-3\sqrt{5}}{2}\right\}\)
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow x^3+6x^2-6x-1=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+x-1=0\)
\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2-\frac{45}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2=\left(\frac{\pm\sqrt{45}}{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\pm\sqrt{45}-7}{2}\end{matrix}\right.\)