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a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)
b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$
c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)
\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)
\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)
$= x + 3 - y$
$= x - y + 3$
(6x3 - 7x2 - x + 2) : (2x + 1)
= (6x3 + 3x2 - 10x2 - 5x + 4x + 2) : (2x + 1)
= [(6x3 + 3x2) - (10x2 + 5x) + (4x + 2)] : (2x + 1)
= [3x2(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)
= (3x2 - 5x + 2)(2x + 1) : (2x + 1)
= 3x2 - 5x + 2
(x4 - x3 + x2 + 3x) : (x2 - 2x + 3)
= (x4 + x3 - 2x3 - 2x2 + 3x2 + 3x) : (x2 - 2x + 3)
= [(x4 + x3) - (2x3 + 2x2) + (3x2 + 3x)] : (x2 - 2x + 3)
= [x3(x + 1) - 2x2(x + 1) + 3x(x + 1)] : (x2 - 2x + 3)
= (x3 - 2x2 + 3x)(x + 1) : (x2 - 2x + 3)
= x(x2 - 2x + 3)(x + 1): (x2 - 2x + 3)
= x(x + 1)
= x2 + x
(x2 - y2 + 6x + 9) : (x + y + 3)
= [(x2 + 6x + 9) - y2] : (x + y + 3)
= [(x + 3)2 - y2] : (x + y + 3)
= (x + 3 + y)(x + 3 - y) : (x + y + 3)
= (x + y + 3)(x - y + 3) : (x + y + 3)
= x - y + 3
CHÚC BN HOK TỐT 
\(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+1\right)\)
\(5\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)\)
\(=\left(a+b\right)\left[5\left(a+b\right)-\left(a-b\right)\right]\)
\(=\left(a+b\right)\left[5a+5b-a+b\right]\)
\(=\left(a+b\right)\left[4a+6b\right]\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
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 :
(*) <=> \(\left(x^2+10+8\right)\left(x^2+10+16\right)\)
a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]
b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2
c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5
1: \(\dfrac{6x^3-7x^2-x+2}{2x+2}\)
\(=\dfrac{6x^3+6x^2-13x^2-13x+12x+12-10}{2x+2}\)
\(=\dfrac{3x^2\left(2x+2\right)-\dfrac{13}{2}x\left(2x+2\right)+6\left(2x+2\right)-10}{2x+2}\)
\(=3x^2-\dfrac{13}{2}x+6-\dfrac{5}{x+1}\)
2: \(\dfrac{x^2-y^2+6x-9}{x+y+3}\)
\(=\dfrac{x^2-\left(y-3\right)^2}{x+y+3}\)
\(=\dfrac{\left(x-y+3\right)\left(x+y-3\right)}{x+y+3}\)