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Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
a) \(\Rightarrow\left(x^2+2\times5x+25\right)+\left(y^2+2y+1\right)\)
\(\Rightarrow\left(x+5\right)^2+\left(y+1\right)^2\)
1)a)x2+10x+26+y2+2y
=(x2+10x+25)+(y2+2y+1)
=(x+5)2+(y+1)2
b)x2-2xy+2y2+2y+1
=(x2-2xy+y2)+(y2+2y+1)
=(x-y)2+(y+1)2
c)z2-6z+13+t2+4t
=(z2-6z+9)+(t2+4t+4)
=(z-3)2+(t+2)2
d)4x2+2z2-4xz-2z+1
=(4x2-4xz+z2)+(z2-2z+1)
=(2x-z)2+(z-1)2
2)a)(x-3)2-4=0
<=>(x-3-2)(x-3+2)=0
<=>(x-5)(x-1)=0
<=>x-5=0 hoặc x-1=0
<=>x=5 hoặc x=1
b)x2-2x=24
<=>x2-2x-24=0
<=>(x2-6x)+(4x-24)=0
<=>x(x-6)+4(x-6)=0
<=>(x-6)(x+4)=0
<=>x-6=0 hoặc x+4=0
<=>x=6 hoặc x=-4
a) x^2 + 10x + 26 + y^2 + 2y
=x2+10x+25+y2+2y+1
=x2+2.x.5+52+y2+2.y.1+12
=(x+5)2+(y+1)2
b)x^2 - 2xy + 2y^2 + 2y +1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
c)z^2 - 6z + 13 + t^2 + 4t
=z2-6z+9+t2+4z+4
=z2-2.z.3+32+t2+2.t.2+22
=(z-3)2+(t+2)2
d)4x^2 + 2z^2 - 4xz - 2z + 1
=4x2-4xz+z2+z2-2z+1
=(2x)2-2.2x.z+z2+z2-2z.1+12
=(2x-z)2+(z-1)2
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(x^2+y^2+26+10x+2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)( do \(\left(x+5\right)^2\ge0;\left(y+1\right)^2\ge0\))
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Bài 1:
a) x( x - y) + x - y = (x - y)(x + 1)
b) 2x + 2y - x( x + y) = ( 2x + 2y) - x( x + y)
= 2( x + y ) - x( x + y ) = ( x + y )(2 - x )
c) 5x2 - 5xy - 10x + 10y = ( 5x2 - 5xy ) - ( 10x - 10y)
= 5x( x - y ) - 10( x - y ) = ( x - y )(5x - 10 )
= 5( x - y )( x - 2 )
d) 4x2 + 6xy - 3x - 6y = Mình ko làm được!!! bạn chép có sai đề không
Bài 2:
x ( 2x - 7) - 4x + 14 = 0
⇒ 2x2 - 7x - 4x + 14 = 0 ⇒ ( 2x2 - 4x ) - ( 7x - 14 ) = 0
⇒ 2x( x - 2 ) - 7(x - 2) = 0
⇒ (x - 2)(2x - 7) = 0
⇒ \(\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x = 2; x = \(\dfrac{7}{2}\)
a) x2+10x+26+y2+2y
=x2+10x+25+y2+2y+1
=(x+5)2+(y+1)2
b) z2-6z+5-t2-4t
=z2-6z+9-t2-4t-4
=(z-3)2-(t2+4t+4)
=(z-3)2-(t+2)2
c)x2-2xy+2y2+2y+1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
d) 4x2-12x-y2+2y+8
=4x2-12x+9-y2+2y-1
=(2x-3)2-(y2-2y+1)
=(2x-3)2-(y-1)2
Đề? ^~~^
\(A=4x^2+4x+8\)
\(=4\left(x^2+x+\dfrac{1}{4}\right)+7\)
\(=4\left(x+\dfrac{1}{2}\right)^2+7\ge7\forall x\)
Vậy Min A = 7 khi \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)
\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5\)
Vậy Min B = 5 khi \(x=0\)
\(C=x^2+10x+26+y^2+2y+2020\)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)+2020\)
\(=\left(x+5\right)^2+\left(y+1\right)^2+2020\ge2020\forall x\)
Vậy Min C = 2020 khi \(\left\{{}\begin{matrix}x+5=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-1\end{matrix}\right.\)