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\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-8x+16=4\)
\(\Leftrightarrow\left(x-2\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)
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...
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
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,\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)
\(\Leftrightarrow3x^2-2x-16=0\)
\(\Leftrightarrow3x^2+6x-8x-16=0\)
\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)
c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b
\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)
\(3x^2-6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
\(x=-\frac{1}{9}\)
\(b.4x^2-12x+9=0\)
\(4x^2-6x-6x+9=0\)
\(2x\left(x-3\right)-3\left(x-3\right)=0\)
\(\left(2x-3\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(x^2+6x+9=\left(x+3\right)^2\)
d) \(4x^2+4x+1=\left(2x+1\right)^2\)
e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)
f) \(4x^2+12x+9=\left(2x+3\right)^2\)
g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)
h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)
a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)2
b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)2
c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)2
d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)2
x²-4=8(x-2)
=> x²-4=8x-16
=> x²-8x+16-4=0
=> (x-4)²-4=0
=>(x-4-2)(x-4+2)=0
=> (x-2)(x-6)=0
=> x-2=0 nên x=2
x-6 =0 nên x=6
a) x^2 - 4 = 8(x - 2)
<=> (x - 2)(x + 2) - 8(x - 2) = 0
<=> (x - 2)(x+2-8)=0
<=>(x-2)(x-6)=0
<=>x-2=0 hoặc x-6=0
<=>x=2 hoặc x=6
Vậy S={2;6}
b)x^2-4x+4=9(x-2)
<=>(x-2)^2-9(x-2)=0
<=>(x-2)(x-2-9)=0
<=>(x-2)(x-11)=0
<=>x-2=0 hoặc x-11=0
<=>x=2 hoặc x=11
Vậy S={2;11}
c)4x^2-12x+9=(5-x)^2
<=>(2x)^2-2.2x.3+3^2=(5-x)^2
<=>(2x-3)^2-(5-x)^2=0
<=>(2x-3-5+x)(2x-3+5-x)=0
<=>(3x-8)(x+2)=0
<=>3x-8=0 hoặc x+2=0
<=>3x=8 hoặc x= - 2
<=>x=8:3(8 phần 3) hoặc x= -2
Vậy S={8:3 ; -2}