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QT
Quoc Tran Anh Le
Giáo viên
1 tháng 7 2019
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow-x^3-6x^2+6x+1=0\)
\(\Leftrightarrow-x^3+x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow-x^2\left(x-1\right)-7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2-7x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-x^2-7x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+7x+12,25-11,25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5\right)^2-\left(\frac{3\sqrt{5}}{2}\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+3,5-\frac{3\sqrt{5}}{2}\right)\left(x+3,5+\frac{3\sqrt{5}}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x+3,5-\frac{3\sqrt{5}}{2}=0\\x+3,5+\frac{3\sqrt{5}}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3,5+\frac{3\sqrt{5}}{2}=\frac{-7+3\sqrt{5}}{2}\\x=-3,5-\frac{3\sqrt{5}}{2}=\frac{-7-3\sqrt{5}}{2}\end{matrix}\right.\)
Vậy x = \(\left\{1;\frac{-7+3\sqrt{5}}{2};\frac{-7-3\sqrt{5}}{2}\right\}\)
1 tháng 7 2019
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow x^3+6x^2-6x-1=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+x-1=0\)
\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2-\frac{45}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{7}{2}\right)^2=\left(\frac{\pm\sqrt{45}}{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\pm\sqrt{45}-7}{2}\end{matrix}\right.\)
DL
1
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
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
LT
13 tháng 8 2020
a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(4-4x\right)=5\)
\(\Leftrightarrow4x-4x^2+8-8x=5\)
\(\Leftrightarrow-4x^2-4x+3=0\)
\(\Leftrightarrow4x^2+4x-3=0\)
\(\Leftrightarrow4x^2-2x+6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)
b) \(6x^2-6x\left(-2+x\right)=36\)
\(\Leftrightarrow6x^2+12x-6x^2=36\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\)
Vậy x = 3
c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)
\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)
\(\Leftrightarrow15-2x=9\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy x = 3
d) \(\left(x+5\right)^2-9=0\)
\(\Leftrightarrow\left(x+5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)
Vậy x ={-2; -8}
e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)
\(\left(x-2\right)^3-x^3+6x^2=7\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)
\(\Leftrightarrow12x-8=7\)
\(\Leftrightarrow12x=15\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
#Chúc bạn học tốt!
\(1+6x-6x^2-x^3=0\)
\(\Leftrightarrow x^2+7x+1-x^3-7x^2-x=0\)
\(\Leftrightarrow\left(x^2+7x+1\right)-x\left(x^2+7x+1\right)=0\)
\(\Leftrightarrow\left(x^2+7x+1\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left[\left(x^2+7x+\frac{49}{4}\right)-\frac{45}{4}\right]\left(1-x\right)=0\)
\(\Leftrightarrow\left[\left(x+\frac{7}{2}\right)^2-\frac{45}{4}\right]\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{45}-7}{2}\\x=-\frac{\sqrt{45}+7}{2}\\x=1\end{matrix}\right.\)