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\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
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
\(5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2=\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]\)
\(\left(y-x\right)^2=\left(x-y\right)^2\)
\(\Rightarrow\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]:\left(y-x\right)^2=5\left(x-y\right)^2-3\left(x-y\right)+4\)
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: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
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: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)