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UN
0
HT
13 tháng 10 2018
\(A=\dfrac{2x^3-18x}{x^4-81}\\ A=\dfrac{2x\left(x^2-9\right)}{\left(x^2+9\right)\left(x^2-9\right)}\\ A=\dfrac{2x}{x^2+9}\)
\(B=\dfrac{x^2-x-20}{x^2-25}\\ B=\dfrac{x\left(x-5\right)+4\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{\left(x-5\right)\left(x+4\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{x+4}{x+5}\)
\(C=\dfrac{8xy-6x^2}{12y^2-9xy}\\ C=\dfrac{2x\left(4y-3x\right)}{3y\left(4y-3x\right)}\\ C=\dfrac{2x}{3y}\)
\(E=\dfrac{x^2+5x+6}{x^2-4}\\ E=\dfrac{x\left(x+2\right)+3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{x+3}{x-2}\)
28 tháng 12 2017
1)Phân tích đa thức thành nhân tử
\(a,x^2+xy+3x+3y\)
\(=x\left(x+y\right)+3\left(x+y\right)\)
\(=\left(x+y\right)\left(x+3\right)\)
\(b,x^2-y^2+4x+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c,x^3+x-y-y^3\)
\(=\left(x^3-y^3\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+1\right)\)
2) \(\dfrac{5}{x+5}-\dfrac{6}{5-x}+\dfrac{x^2+25}{x^2-25}\)
\(=\dfrac{5}{x+5}+\dfrac{6}{x-5}+\dfrac{x^2+25}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{5\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{6\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{x^2+25}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{5x-25+6x+30+x^2+25}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+11x+30}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+5x+6x+30}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x+6}{x-5}\)
\(3,\dfrac{x}{x^2-4}+\dfrac{2}{x-2}+\dfrac{2}{x+2}\)
\(=\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2x+4+2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{5x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x^2-4}\)
S
13 tháng 9 2020
\(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(x^2+10x+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
\(b,x-y+5=a\text{ thì biểu thức bằng:}a^2-2a+4>0\text{ nên k phân tích đc}\)
\(d,x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
13 tháng 9 2020
a) 100x2 - ( x2 + 25 )2
= ( 10x )2 - ( x2 + 25 )2
= [ 10x - ( x2 + 25 ) ][ 10x + ( x2 + 25 ) ]
= ( -x2 + 10x - 25 )( x2 + 10x + 25 )
= -( x2 - 10x + 25 )( x2 + 10x + 25 )
= -( x - 5 )2( x + 5 )2
b) ( x - y + 5 )2 + 4 - 4( x - y + 5 ) ( 4 may ra còn phân tích được :)) )
= ( x - y + 5 )2 - 2( x - y + 5 ).2 + 22
= ( x - y + 5 - 2 )2
= ( x - y + 3 )2
c) a2 - 25( b - c ) ( không phân tích được :)) )
d) x3 + 27y3 = x3 + ( 3y )3 = ( x + 3y )( x2 - 3xy + 9y2 )
CN
1
14 tháng 3 2020
a) x2 + 10x + 25 - 4x2 - 20x = 0
<=> 3x2 + 10x - 25 = 0
<=> (x + 5)(3x - 5) = 0 <=> \(\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)
Vậy S = \(\left\{-5;\frac{5}{3}\right\}\)
b. (4x - 5)2 - 2(4x - 5)(4x + 5) = 0
<=> (4x - 5)[(4x - 5) - 2(4x + 5)] = 0
<=> (4x - 5)(4x - 5 - 8x - 10) = 0
<=> (4x - 5)(-4x - 15) = 0 <=> \(\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}\)
Vậy S = \(\left\{-\frac{15}{4};\frac{5}{4}\right\}\)
20 tháng 10 2022
a: \(=\left(a+b+c\right)^2+2\left(a+b+c\right)\left(b-c\right)+\left(b-c\right)^2\)
\(=\left(a+b+c+b-c\right)^2=\left(a+2b\right)^2\)
b: \(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(=\left(x^4+4\right)\left(x^4-4\right)=x^8-16\)
d: \(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)-3\left(x^2+y^2+z^2\right)\)
=0
HH
21 tháng 6 2018
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
PL
2 tháng 8 2018
a)Ta có : 9(a + b)2 - 4(a - 2b)2
= [3(a + b) - 2(a - 2b)].[3(a + b) + 2(a - 2b)]
= (3a + 3b - 2a + 4b)(3a + 3b + 2a - 4b)
= (a + 7b)(5a - b)
\(-x^2+25=5^2-x^2=\left(5-x\right)\left(5+x\right)\)
\(25-x^2=\left(5-x\right)\left(x+5\right)\)