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1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
B = \(\frac{8xy-6x^2}{3y\left(3x-4y\right)}=\frac{2x\left(4y-3x\right)}{-3y\left(4y-3x\right)}=-\frac{2x}{3y}\)
C = \(\frac{2x^3-18x}{x^4-81}=\frac{2x\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\frac{2x}{x^2+9}\)
\(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2\)
\(=3xy\left(x+y\right)\)
\(a,\left(2x+3y\right)^2-4\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x+3y-4\right)\)
\(b,\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]\)
\(=\left(x+y\right).3x\)
\(c,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
a) \(\left(3x^2y-11x^2-5y\right)\left(8xy-5x+6\right)\)
\(=3x^2y\left(8xy-5x+6\right)-11x^2\left(8xy-5x+6\right)-5y\left(8xy-5x+6\right)\)
\(=24x^3y^2-15x^3y+18x^2y-88x^3y+55x^3-66x^2-40xy^2+25xy-30y\)
\(=24x^3y^2-103x^3y+18x^2y+55x^3-66x^2-40xy^2+25xy-30y\)
b) \(\left(-4x^2y-5x^2+3y^3\right)\left(2x^2-xy+3y^2\right)\)
\(=-4x^2y\left(2x^2-xy+3y^2\right)-5x^2\left(2x^2-xy+3y^2\right)+3y^3\left(2x^2-xy+3y^2\right)\)
\(=-8x^4y+4x^3y^2-12x^2y^3-10x^4+5x^3y-15x^2y^2+6x^2y^3-3xy^4+9y^5\)
\(=-8x^4y+4x^3y^2-6x^2y^3-10x^4+5x^3y-15x^2y^2-3xy^4+9y^5\)
P/s: Ko chắc ạ!
D= 5x^2+8xy+5y^2-2x+2y
=4x^2+8xy+4y^2-2x+2y+y^2+x^2
=(2x+2y)^2+x^2-2*1/2x+1/4+y^2+2*1/2y+1/4-1/2
(2x+2y)^2+(x-1/2)^2+(y+1/2)^2-1/2>=-1/2
suy ra D>=-1/2 nên D có GTNN là -1/2
Ta có : 5D = 25x2 + 40xy + 25y2 - 10x + 10y
5D = (5x+ 4y - 1)2 + 9y2 + 18y - 1
5D = ( 5x + 4y - 1)2 + 9 (y + 1)2 - 2
D =\(\frac{1}{5}\). ( 5x + 4y - 1)2 + \(\frac{9}{5}\).( y + 1)2 - \(\frac{2}{5}\) \(\ge\)\(\frac{-2}{5}\)
Dấu "=" xảy ra khi y+1 = 0 \(\Leftrightarrow\)y = -1
5x + 4y - 1 = 0 \(\Leftrightarrow\)x=1
Vậy GTNN của D = \(\frac{-2}{5}\)khi x = 1 ; y = -1
Bạn dùng hằng đẳng thức ý
a. (A – B)2= A2 – 2AB+ B2
b. (A – B)2= A2 – 2AB+ B2
c. (2x+3y)(2x-3y)-(2x+y)2
= 4x2-9y2 -4x2 - 4xy-y2
= -10y2-4x2
d. (A+B)2 = A2+2AB+B2
=-4x^2-9y^2-4(x^2-2xy+y^2)-8xy
=4x^2-9y^2-4x^2+8xy-4y^2-8xy
= -13y^2
đề bài như thế nào bn