a) (x² +5x): x
b) (6x3-4x²): (2x²)
c) (8x²y + 6xy): (2xy)
d) (x² - x + 3). (x-2)
e) (8x³y): (4xy)
f) (-4x y²). (- 1/4 x)
g) (x4 - 2x² +1): (x-1)
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Để tính các biểu thức trên, ta sẽ áp dụng quy tắc nhân đa thức.
a) 2xy(3x+1) = 6x^2y + 2xy
b) -6x^2y(4x-5) = -24x^3y + 30x^2y
c) -3x^2(4x^2y-6xy) = -12x^4y + 18x^3y
d) 1/2xy^2(2x+3) = xy^2 + 3/2xy^2
e) 8x^2y^2(1/4xy-1/2x^2) = 2xy - 4x^2y^2
f) 5x(x^2+3x+1) = 5x^3 + 15x^2 + 5x
g) -1/2x^2y(2xy+6) = -x^3y - 3x^2y
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
a: \(=7x\left(xy-3\right)\)
d: \(=\left(x+1\right)\left(10x-8y\right)\)
\(=2\left(5x-4y\right)\left(x+1\right)\)
e: \(=\left(x-100\right)\cdot7x\)
f: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
mình k cho bạn rồi nha, tích lại cho mình, số điểm của mình là -159 điểm
a: \(\dfrac{x^2+5x}{x}=\dfrac{x^2}{x}+\dfrac{5x}{x}=x+5\)
b: \(\dfrac{6x^3-4x^2}{2x^2}=\dfrac{6x^3}{2x^2}-\dfrac{4x^2}{2x^2}=3x-2\)
c: \(\dfrac{8x^2y+6xy}{2xy}=\dfrac{8x^2y}{2xy}+\dfrac{6xy}{2xy}=4x+3\)
d: \(\left(x^2-x+3\right)\left(x-2\right)\)
\(=x^3-2x^2-x^2+2x+3x-6\)
\(=x^3-3x^2+5x-6\)
e: \(\dfrac{8x^3y}{4xy}=\dfrac{8}{4}\cdot\dfrac{x^3}{x}\cdot\dfrac{y}{y}=2x^2\)
f: \(\left(-4xy^2\right)\cdot\left(-\dfrac{1}{4}x\right)=\left(-4\right)\cdot\left(-\dfrac{1}{4}\right)\cdot x\cdot x\cdot y^2=x^2y^2\)
g: \(\dfrac{x^4-2x^2+1}{x-1}\)
\(=\dfrac{\left(x^2-1\right)^2}{x-1}=\dfrac{\left[\left(x-1\right)\left(x+1\right)\right]^2}{\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x-1\right)}=\left(x-1\right)\left(x+1\right)^2\)
Lời giải:
a. $(x^2+5x):x=x(x+5):x=x+5$
b. $(6x^4-4x^2):(2x^2)=2x^2(3x^2-2):(2x^2)=3x^2-2$
c. $(8x^2y+6xy):(2xy)=2xy(4x+3):(2xy)=4x+3$
d.
$(x^2-x+3)(x-2)=x(x^2-x+3)-2(x^2-x+3)$
$=x^3-x^2+3x-2x^2+2x-6=x^3-3x^2+5x-6$
e.
$(8x^3y):(4xy)=2x^2$
f.
$(-4xy^2).\frac{-1}{4}x=x^2y^2$
g.
$(x^4-2x^2+1):(x-1)=(x^2-1)^2:(x-1)=(x-1)^2(x+1)^2:(x-1)=(x-1)(x+1)^2$