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TN
8 tháng 5 2018
a) Ta có: \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-5}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)\(\Rightarrow x-5=3x-11\Rightarrow x-3x=-11+5\Rightarrow-2x=-6\Rightarrow x=3\)
b)Ta có: \(\frac{15-6x}{3}>5\)
\(\Rightarrow15-6x>15\)
\(\Rightarrow6x< 0\)
\(\Rightarrow x< 0\).
Kb với mình nha!
3 tháng 12 2015
\(A=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{\left(x-3\right)\left(x+3\right)}=\frac{3\left(x-3\right)+\left(x+3\right)+18}{\left(x-3\right)\left(x+3\right)}=\frac{4x+12}{\left(x-3\right)\left(x+3\right)}=\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)
Với x = 1
\(A=\frac{4}{x-3}=\frac{4}{1-3}=\frac{4}{-2}=-2\)
5 tháng 8 2021
a) = 5( x2 - 9y2 - 6y - 1 ) = 5[ x2 - ( 9y2 + 6y + 1 ) ] = 5[ x2 - ( 3y + 1 )2 ] = 5( x - 3y - 1 )( x + 3y + 1 )
b) = 125x3 - 25x2 + 15x2 - 3x + 5x - 1 = 25x2( 5x - 1 ) + 3x( 5x - 1 ) + ( 5x - 1 ) = ( 5x - 1 )( 25x2 + 3x + 1 )
c) = 5( x - 7 ) + a( x - 7 ) = ( x - 7 )( a + 5 )
d) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
e) = ax2 + a - a2x - x = ax( a - x ) + ( a - x ) = ( a - x )( ax + 1 )
f) = ( 10x )2 - ( x2 + 25 )2 = ( 10x - x2 - 25 )( 10x + x2 + 25 ) = -( x - 5 )2( x + 5 )2
28 tháng 11 2021
g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
\(\left\{{}\begin{matrix}\left|x\right|\ge2\\\left|y\right|\ge2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2\ge4\\y^2\ge4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}\le\dfrac{1}{4}\\\dfrac{1}{y^2}\le\dfrac{1}{4}\end{matrix}\right.\)
\(\left(\dfrac{x+y}{xy}\right)^2=\dfrac{\left(x+y\right)^2}{x^2y^2}\le\dfrac{2\left(x^2+y^2\right)}{x^2y^2}=\dfrac{2}{x^2}+\dfrac{2}{y^2}\le2.\dfrac{1}{4}+2.\dfrac{1}{4}=1\)
\(\Rightarrow\dfrac{x+y}{xy}\le1\)
Dấu "=" xảy ra khi \(x=y=\pm2\)