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a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
a) 12xy( x2 - 2xy + y2) = 12xy( x - y )2
b) ( x2 + xy ) - ( 6x + 6y ) = x( x + y ) - 6( x + y )
= ( x + y )(x - 6)
c) ( 2x2 + 2xy ) - ( x + y ) = 2x(x + y ) - ( x + y )
= (x + y )(2x - 1)
e) ( 3x2 - 3y2 ) - ( 12x + 12y ) = 3( x2 - y2 ) - 12( x + y)
= 3(x - y)(x + y) - 12(x + y) = ( x + y )(3x - 3y - 12)
= 3( x + y )(x - y -4)
g) \(\left[x\left(x+10\right)\right].\left[\left(x+4\right)\left(x+6\right)\right]\) + 128
= (x2 + 10x).(x2 + 10x + 24) + 128
Đặt x2 + 10x + 12 = t
⇒ Biểu thức trên có dạng:
( t - 12 )(t + 12) + 128 = t2 - 144 + 128 = t2 - 16 = t2 - 42
= ( t - 4 )( t + 4) = (x2 + 10x + 12 - 4 )(x2 + 10x + 12 + 4)
= ( x2 + 10x + 8)(x2 + 10x + 16)
f) -2xy + 4y2 = 2y( -x + 2y)
Có 2 phần g nha bạn. Mk chuyển phần cuối thành phần f.
Phần d do mk hơi ngu nên chưa nghĩ ra bạn thông cảm nha.
a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
1 , \(x^5+x^4+1=\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
= \(x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)=\(\left(x^2+x+1\right)\left(x^3-x+1\right)\)
2 , \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)(*)
Đặt x2 + 10 = a , a>0 (1)
=> (*) <=> a(a+24)+128=a2 + 24a+128=(a+8)(a+16) (**)
Thay (1) vào (**) ta được :
(*) <=> \(\left(x^2+10+8\right)\left(x^2+10+16\right)\)