Bài 1: Tìm x
a) (x-1)^3+3(x+1)^2=(x^2-2x+4)(x+2)
b) x^2-4=8(x-2)
c) x^2-4x+4=9(x-2)
d) 4x^2-12x+9=(5-x)^2
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HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
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
6 tháng 12 2019
\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)
\(3x^2-6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
\(x=-\frac{1}{9}\)
\(b.4x^2-12x+9=0\)
\(4x^2-6x-6x+9=0\)
\(2x\left(x-3\right)-3\left(x-3\right)=0\)
\(\left(2x-3\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
6 tháng 12 2019
\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
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,\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)
\(\Leftrightarrow3x^2-2x-16=0\)
\(\Leftrightarrow3x^2+6x-8x-16=0\)
\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)
\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-8x+16=4\)
\(\Leftrightarrow\left(x-2\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)
Vậy...
\(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow\left(x+\frac{13}{2}\right)^2=\frac{81}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-21}{2}\end{cases}}\)
Vậy...