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PA
1
26 tháng 11 2019
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 = \(\frac{2x^3-18x}{x^4-81}=\frac{2x\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\frac{2x}{x^2+9}\)
12 tháng 7 2022
a: \(=3\left(x^2-\dfrac{2}{3}x+\dfrac{4}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{11}{3}>=\dfrac{11}{3}\)
Dấu '=' xảy ra khi x=1/3
b: \(=2\left(x^2+\dfrac{3}{2}x\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}\right)\)
\(=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{9}{8}>=-\dfrac{9}{8}\)
Dấu '=' xảy ra khi x=-3/4
d: \(=3\left(x^2-2x+\dfrac{2}{3}\right)\)
\(=3\left(x^2-2x+1-\dfrac{1}{3}\right)\)
\(=3\left(x-1\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=1
NT
1
24 tháng 9 2017
\(a,x^3-3x^2+3x-9=0\\ \Leftrightarrow x^2\left(x-3\right)+3\left(x-3\right)=0\\\Leftrightarrow \left(x-3\right)\left(x^2+3\right)=0\\ \Leftrightarrow x-3=0\left(dox^2+3\ge3>0\right)\\ \Leftrightarrow x=3\)
Vậy...
\(b,x^2+3y^2+2xy+4y+2x+3=0\\ \Leftrightarrow\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+\left(2y^2+2y+2\right)=0\\ \Leftrightarrow\left(x+y+1\right)^2+2\left[\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
Với mọi x;y thì \(\left(x+y+1\right)^2\ge0\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)
\(\Rightarrow\left(x+y+1\right)^2+2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Do đó ko tìm đc gtri nào củax;y thoa mãn
S
1
DM
3 tháng 11 2018
a) x2+3x+2-x3-27
b) (x2-6x+9)+(4y2+4y+1)=0
( x-3)2+ (2y+1)=0 => (x-3)2=0 hoặc (2y+1)=0
=> x=3, y= -1/2
c)152+852+2.15.85+5100= (15+85)2+5100
PN
0
19 tháng 10 2020
a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
NN
19 tháng 10 2020
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)