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Ta có: \(\left(\dfrac{x^2-3x}{x^2-9}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)
\(=\left(\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right):\left(\dfrac{9-x^2+x^2-9+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\right)\)
\(=\left(\dfrac{x}{x+3}-1\right):\dfrac{x-2}{x+3}\)
\(=\dfrac{x-x-3}{x+3}\cdot\dfrac{x+3}{x-2}\)
\(=\dfrac{-3}{x-2}\)
Điều kiện : x ≠ 2 ; x ≠ 3 ; x ≠ - 3
\(\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-1\right):\left(\dfrac{\left(3-x\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)
\(=\left(\dfrac{x}{x+3}-1\right):\left(\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\right)\)
\(=\dfrac{x-x-3}{x+3}:\dfrac{9-x^2+x^2-9+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+3}:\dfrac{x-2}{\left(x+3\right)}\)
\(=\dfrac{-3}{x-2}\)
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
Phân tích thôi hả :v
( 3x - 7 )2 - 25 = ( 3x - 7 )2 - 52 = ( 3x - 7 - 5 )( 3x - 7 + 5 ) = ( 3x - 12 )( 3x - 2 ) = 3( x - 4 )( 3x - 2 )
( x - 1/2 )2 - 9/4 = ( x - 1/2 )2 - ( 3/2 )2 = ( x - 1/2 - 3/2 )( x - 1/2 + 3/2 ) = ( x - 2 )( x + 1 )
49 - ( x + 7 )2 = 72 - ( x + 7 )2 = [ 7 - ( x + 7 ) ][ 7 + ( x + 7 ) ] = ( 7 - x - 7 )( 7 + x + 7 ) = -x( x + 14 )
25 - ( x - 3 )2 = 52 - ( x - 3 )2 = [ 5 - ( x - 3 ) ][ 5 + ( x - 3 ) ] = ( 5 - x + 3 )( 5 + x - 3 ) = ( 8 - x )( x + 2 )
không ảnh hưởng
\(\dfrac{\left(x-9\right)^3}{2\left(9-x\right)}=\dfrac{-\left(x-9\right)^3}{-2\left(9-x\right)}=\dfrac{-\left(x-9\right)^3}{2\left(x-9\right)}=\dfrac{\left(x-9\right)^3}{-2\left(x-9\right)}\)
\(1,\left(2-x\right)^2-9=0\)
\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)
\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)
1. \(\left(2-x\right)^2-9=0\)
\(\left(2-x\right)^2=9\)
\(\left(2-x\right)^2=3^2\)
\(2-x=3\)
\(-x=-1\Rightarrow x=1\)