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24 tháng 8 2018
a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
Vậy \(S=\left\{\dfrac{8}{5}\right\}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-6=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-6=0\Leftrightarrow x=6\)
Vậy \(S=\left\{2;6\right\}\)
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-11=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-11=0\Leftrightarrow x=11\)
Vậy \(S=\left\{2;11\right\}\)
(d ko bít lèm)
#IDOL
4 tháng 8 2019
a,\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x-1+3x^2+6x+3=8\)
\(\Leftrightarrow9x=6\)
\(\Leftrightarrow x=\frac{2}{3}\)
b,\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2+12x-8x=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
c,\(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow x^2-4x+4=9x-18\)
\(\Leftrightarrow x^2-4x+4-9x+18=0\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow x^2-2x-11x+22=0\)
\(\Leftrightarrow x\left(x-2\right)-11\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
d,\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)
\(\Leftrightarrow3x^2-2x-16=0\)
\(\Leftrightarrow3x^2+6x-8x-16=0\)
\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)
10 tháng 1 2017
Bài 1:
\(x^4-4x^3+12x-9=0\)
\(\Rightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Rightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Rightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left(x^2-3x-x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[x\left(x-3\right)-\left(x-3\right)\right]\left(x^2-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)
Bài 2:
\(x^4-4x^3+3x^2+4x-4=0\)
\(\Rightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Rightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x^2-4x+4\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)^2\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
17 tháng 7 2017
a ) 4x^2 + 12x + 9
= ( 2x )^2 + 2.2x.3 + 3^2
= ( 2x + 3 ) ^2
a)x4-4x3+12x-9 = x3(x-1) -3x2(x-1) -3x(x-1) +9(x-1)
=(x-1)(x3-3x2-3x+9)
=(x-1)[x2(x-3)-3(x-3)]
=(x-1)(x-3)(x2-3)
b)(x+2)2=9(x2-4x+4) <--> x2+4x+4=9x2-36x+36
<-->8x2 -40x+32=0
<-->8(x2-5x+4)=0
<-->x2-5x+4=0
<--->(x-4)(x-1)=0
* Nếu x-4=0 <--> x=4
* Nếu x-1=0<--> x=1
Vậy S={4;1}