A-(8x^2–12xy)=9x^2+xy
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a. \(9x^2+25-12xy+5y^2-10y\)
\(=\left(9x^2-12xy+4y^2\right)+\left(25+y^2-10y\right)\)
\(=9\left(x^2-\frac{4xy}{3}+\frac{4y^2}{9}\right)+\left(5-y\right)^2\)
\(=9\left(x-\frac{2y}{3}\right)^2+\left(5-y\right)^2\)
\(a,x^3+xy-2y-8\)
\(=\left(x^3-8\right)+\left(xy-2y\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4+y\right)\)
\(b,8x^3-12xy+2x^2y-3y^2\)
\(=\left(8x^3+2x^2y\right)-\left(12xy+3y^2\right)\)
\(=2x^2\left(4x+y\right)-3y\left(4x+y\right)\)
\(=\left(2x^2-3y\right)\left(4x+y\right)\)
a) 9x2 + 25 - 12xy + 5y2 - 10y
= ( 9x2 - 12xy + 4y2 ) + ( y2 - 10y + 25 )
= ( 3x - 2y )2 + ( y - 5 )2
b) 13x2 + 4x + 12xy + 4y2 + 1
= ( 9x2 + 12xy + 4y2 ) + ( 4x2 + 4x + 1 )
= ( 3x + 2y )2 + ( 2x + 1 )2
c) x2 + 20 + 9y2 + 8x - 12y
= ( x2 + 8x + 16 ) + ( 9y2 - 12y + 4 )
= ( x + 4 )2 + ( 3y - 2 )2
a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
a: =6xy+xy=7xy
b: =-9xy^2
c: =-x^2y^3z^4
d: =-4x^2y
e: =-30x^2y
f: =6x^2y
a)x2-4y2
=x2-(2y)2
=(x-2y)(x+2y)
b)x3+27y3
=x3+(3y)3
=(x+3y)(x2-3xy+9y2)
c)4x2+12xy+9y2-16
=(2x+3y)2-42
=(2x+3y-4)(2x+3y+4)
d)9x2-24xy+16y2-64
=(3x-4y)2-82
=(3x-4y-8)(3x-4y+8)
e)8x3-27y3
=(2x)3-(3y)3
=(2x-3y)(4x2+6xy+9y2)
f)5x3-7x2+10x-14
=5x3+10x-7x2-14
=5x(x2+2)-7(x2+2)
=(5x-7)(x2+2)
Bài 2: Tìm x
a) x2 - 6x + 5 = 0
<=> x2 - x - 5x + 5 = 0
<=> x(x - 1) - 5(x - 1) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
b) x2 - 2x - 24 = 0
<=> x2 + 4x - 6x - 24 = 0
<=> x(x + 4) - 6(x + 4) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
Vậy x ={-4; 6}
\(-\left(8x^2-12xy\right)=9x^2+xy\)
\(\Leftrightarrow-8x^2+12xy=9x^2+xy\)
\(\Leftrightarrow17x^2-11xy=0\)
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