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c.
C=6(xy)^2-6(xy)y^2-(2x)^3+8(xy)^2+5(xy)^2-5(xy).y^2
C=(6+8+5)(xy)^2-(6+5)(xy)^2.y^2 -(2x)^3+8.(xy)^2
x.y=1; 2x=1
C=19-11.4-1+8
C=26-44=30-40-4-4=-10-8=-18
a)
<=>A=3x[10x^2-2x+1-2(5x^2-x-2)]=3x(1+4)
=3.5.x
x=15
A=3.5.15=15^2=(4^2-1).15=4.15.4-15=60.4-15
=240-15=225
\(1,8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\\ 3,x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)\\ =\left(x-y\right)\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(x-y-1\right)\\ 4,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2\\ =\left(x-y-z\right)\left(x-y+z\right)\\ 5,x^2-3x+xy-3y=x\left(x-3\right)+y\left(x-3\right)\\ =\left(x-3\right)\left(x+y\right)\)
\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
b, đề thiếu nhé
\(c,x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
\(d,x^2-2xy+y^2-z^2=\left(x+y\right)^2-z^2\)
\(=\left(x+y-z\right)\left(x+y+z\right)\)
\(e,x^2-3x+xy-3y=\left(x^2-3x\right)+\left(xy-3y\right)\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
câu g hình như sai đề rồi đó ❤ NTN ❤
Khôi Bùi Ông hok rồi ông giúp tui câu b và g cái coi
a, 7x^3 + 5 ( x - y )^2 v- 7y^3
= 7 ( x^3 - y^3 ) + 5 ( x-y )^2
= 7 ( x - y )^3 + 5 ( x-y ) ^2
= [ 7 ( x- y ) + 5 ] ( x-y) ^2
c) \(x^2+x-ax-a\)
\(=x\left(x+1\right)-a\left(x+1\right)\)
\(=\left(x+1\right)\left(x-a\right)\)
d) \(2xy-ax+x^2-2ay\)
\(=2y\left(x-a\right)+x\left(x-a\right)\)
\(=\left(x-a\right)\left(2y+x\right)\)
e) \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
f) \(25-10x-4y^2+x^2\)
\(=\left(x^2-10x+25\right)-\left(2y\right)^2\)
\(=\left(x-5\right)^2-\left(2y\right)^2\)
\(=\left(x-5-2y\right)\left(x-5+2y\right)\)
g) \(x^3-6xy+9y^2-36\)
h) \(4x^2-9y^2+4x-6y\)
\(=\left(2x\right)^2-\left(3y\right)^2+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
k) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(-x+y+5\right)\)
i) \(4x^2-25y^2-6x+15y\)
\(=\left(2x\right)^2-\left(5y\right)^2-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y\right)-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y-3\right)\)
a, \(x\left(y+z\right)^2+y\left(x+z\right)^2+z\left(x+y\right)^2+4xyz\)
\(=x\left(y+z\right)^2+x^2\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+xz+z^2+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+xz^2-x^2z-x^2y-xy^2\)
\(=yz\left(y+z\right)-x\left(y+z\right)\left(y-z\right)-x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(yz-xy+xz-x^2\right)\)
\(=\left(y+z\right)\left[y\left(z-x\right)+x\left(z-x\right)\right]\)
\(=\left(y+z\right)\left(y+x\right)\left(z-x\right)\)
a, A=2x2+y2-2xy-2x+3
= (x2-2xy+y2)+(2x2-2x+2)+1
=(x-y)2+2(x-1)2+1
vì (x-y)2 ≥0 ∀x,y
(x-1)2 ≥ 0 ∀x
=> (x-y)2+2(x-1)2+1 ≥1 ∀x,y
=> A ≥1
= > GTNN A = 1 khi
x-1=0
=> x=1
x-y=0
=> 1-y=0
=> y=1
vậy GTNN A =1 khi x=y=1
\(6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2+xy\right)\\ =6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2+5xy^3\\ =\left(6x^2y^2+8x^2y^2+5x^2y^2\right)+\left(-6xy^3+5xy^3\right)-8x^3\\ =19x^2y^2-xy^3-8x^3\)
Với `x=1/2;y=2` ta có :
\(19x^2y^2-xy^3-8x^3\\ =19.\left(\dfrac{1}{2}\right)^2.2^2-\dfrac{1}{2}.2^3-8.2^3\\ =19.\dfrac{1}{4}.4-\dfrac{1}{2}.8-8.8\\ =19-4-64\\ =-49\)