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1/ x^2 +4xy +4y^2 = (x +2y)^2
2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3
3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3
4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3
\(x^2+4y^2+13-6x+8y=0\)
\(=\left(x-3\right)^2+4\left(y-1\right)^2-26\ge-26\)
\(Min\) là \(-26\Leftrightarrow x=3;y=1\)
Vậy................
bài 3:
b) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow x^2-2x+1+y^2-4y+4=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy x=1; y=2
c) \(x^2+4y^2+13-6x-8y=0\)
\(\Leftrightarrow x^2-6x+9+4y^2-8y+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vây x=3; y=1
Bài 3:
a) \(x\left(x+4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow x^2+4x-5x+20=0\)
\(\Leftrightarrow x^2-x+20=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+20=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{79}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{-79}{4}\)
\(\Rightarrow\) ptvn
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
2 .tìm x
a , x ( x + 2 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b, x ( x-5 )= 5 -x
<=> x ( x-5 ) + x - 5 = 0
<=> x (x-5) + ( x-5)= 0
<=> (x-5)(x+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
c) ( x + 1 ) ( 6x2 + 2x ) + ( x - 1 ) ( 6x2 + 2x ) = 0
\(\Leftrightarrow\) ( 6x2 + 2x ) \([\)(x+1)(x-1)\(]\)=0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x
= x ( 2a + 1 ) - y ( 2a + 1 )
= ( 2a + 1 ) ( x - y )
b) a2 ( x - y ) - ( y - x ) = a2x - a2y - y + x
= x ( a2+ 1 ) - y ( a2 +1 )
= ( a2+1 ) - (x-y )
c) x ( x - y ) + y ( y - x ) - 3 ( x - y ) = x 2 - xy -+ y 2 - xy - 3x + 3y
= x2 - 2xy + y2 -3x + 3y
= (x-y)2 -3 ( x - y )
= ( x-y ) ( x-y+3)
a) \(x^2+4y^2-6x-4y+10=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\2y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)
b) \(2x^2+y^2+2xy-10x+25=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x-5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-5\\x=5\end{cases}}\)
c) \(x^2+2xy+4x-4y-2xy+5=0\)
\(\Leftrightarrow x^2-4x-4y+5=0\)
Xem lại đề câu c).
a) x2 + 4y2 - 6x - 4y + 10 = 0
<=> x2 - 6x + 9 + 4y2 - 4y + 1 = 0
<=> ( x - 3 )2 + ( 4y - 1 )2 = 0
<=> \(\hept{\begin{cases}x-3=0\\4y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{4}\end{cases}}\)
b) 2x2 + y2 + 2xy - 10x + 25 = 0
<=> x2 + 2xy + y2 + x2 - 10x + 25 = 0
<=> ( x + y )2 + ( x - 5 )2 = 0
<=> \(\hept{\begin{cases}x+y=0\\x-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5\\x=5\end{cases}}\)
c) Xem lại đề
\(x^2-2x+5+y^2-4y=0\)
\(x^2-2\times x\times1+1^2-1^2+y^2-2\times y\times2+2^2-2^2+5=0\)
\(\left(x-1\right)^2+\left(y-2\right)^2=0\)
\(\left(x-1\right)^2\ge0\)
\(\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2=\left(y-2\right)^2=0\)
\(\Leftrightarrow x-1=y-2=0\)
\(\Leftrightarrow x=1;y=2\)
\(x^2+4y^2+13-6x-8y=0\)
\(\Leftrightarrow x^2-6x+9+4y^2-8y+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-2\right)^2=0\)
Dấu = xảy ra khi
\(\orbr{\begin{cases}x-3=0\\2y-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\y=1\end{cases}}\)