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1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
a,x^4+2x^3-4x-4
=(x^3+2x^3)-(4x+4)
=x^3(x+2)-4(x+2)
=(x^3-4)(x+2)
\(X^4+2X^3-4X-4\)
\(=\left(X^2\right)^2+2X^3-4X-2^2\)
\(=\left[\left(X^2\right)^2-2^2\right]+\left[2X^3-4X\right]\)
\(=\left(X^2+2\right)\left(X^2-2\right)+2X\left(X^2-2\right)\)
\(=\left(X^2-2\right)\left(X^2+2+2X\right)\)
a) -x2 + 2x - 1
= -( x2 - 2x + 1 )
= -( x - 1 )2
b) 12y - 36 - y2
= -( y2 - 12y + 36 )
= -( y - 6 )2
c) -x3 + 9x2 - 27x + 27
= -( x3 - 9x2 + 27x - 27 )
= -( x - 3 )3
d) x3 - 6x2 + 9x
= x( x2 - 6x + 9 )
= x( x - 3 )2
e) a3b - ab3
= ab( a2 - b2 )
= ab( a - b )( a + b )
f) a2 + 2a + 1 - b2
= a2 + ab + a - ab - b2 - b + a + b + 1
= a( a + b + 1 ) - b( a + b + 1 ) + 1( a + b + 1 )
= ( a - b + 1 )( a + b + 1 )
a)\(-x^2+2x-1\)
\(=-\left(x^2-2x+1\right)\)
\(=-\left(x-1\right)^2\)
b) \(12y-36-y^2\)
\(=-\left(y^2-12y+36\right)\)
\(=-\left(y^2-2\cdot1\cdot6+6^2\right)\)
\(=-\left(y-6\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(-x^2+6x-9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-6x+9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=-\left(x-3\right)\left(x-3\right)^2\)
\(=\left(x-3\right)^3\)
d) \(x^3-6x^2+9\)
\(=x\left(x^2-6x+9\right)\)
\(=x\left(x-3\right)^2\)
e) \(a^3b-ab^3\)
\(=ab\left(a^2-b^2\right)\)
\(=ab\left(a-b\right)\left(a+b\right)\)
f) \(a^2+2a+1-b^2\)
\(=a^2+2\cdot a\cdot1+1^2-b^2\)
\(=\left(a+1\right)^2-b^2\)
\(=\left(a+1-b\right)\left(a+1+b\right)\)
a) x2 + 4x + 3 - y2 -2y
= x2 +4x + 4 - y2 -2y-1
= (x+2)2 - (y+1)2
= (x+2-y-1).(x+2+y+1)
= (x-y+1).(x+y+3)
b) 2a2 -5ab + 2b2
= 2a2 -4ab + 2b2 - ab
= 2.(a2 - 2ab+b2) - ab
= 2.(a-b)2 -ab
...
c) (x+y)2 - 2x - 2y + 1
= (x+y)2 - 1 - 2x -2y +2
= (x+y-1).(x+y+1) - 2.(x+y-1)
= (x+y-1)2
1
a) x2 + 4y2 + 4xy - 16
=(x2 + 4xy + 4y2) - 16
=(x+2y)2 - 16
=(x+2y-4)(x+2y+4)
b)x2 + y2 - 2x + 4y + 5 =0
<=> x2 - 2x + 1 + y2 - 4y + 4=0
<=> (x-1)2 + (y-2)2 =0
<=> x=1 và y=2
1)
a) \(2x^2-12x+18+2xy-6y\)
\(=2x^2-6x-6x+18+2xy-6y\)
\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)
\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)
\(=\left(x-3\right)\left(2y+2x-6\right)\)
\(=2\left(x-3\right)\left(y+x-3\right)\)
b) \(x^2+4x-4y^2+8y\)
\(=x^2+4x-4y^2+8y+2xy-2xy\)
\(=\left(-4y^2+2xy+8y\right)+\left(-2xy+x^2+4x\right)\)
\(=2y\left(-2y+x+4\right)+x\left(-2y+x+4\right)\)
\(=\left(2y+x\right)\left(-2y+x+4\right)\)
2) \(5x^3-3x^2+10x-6=0\)
\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)
Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)
\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)
\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
Bài làm
a) 2x2 - 12x + 18 + 2xy - 6y
= 2x2 - 6x - 6x + 18 + 2xy - 6y
= ( 2xy + 2x2 - 6x ) - ( 6y + 6x - 18 )
= 2x( y + x - 3 ) - 6( y + x - 3 )
= ( 2x - 6 ) ( y + x - 3 )
# Học tốt #
a) \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
b) \(x^4+2x^2y+y^2\)
\(=\left(x^2+y\right)^2\)
\(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(\left(3x-5\right)^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\)
\(\Leftrightarrow x=3,x=1\)
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