15x/x2 +3x -4 -1= 12(1/x+4+1/3x-3
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8 tháng 9 2023
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
26 tháng 7 2017
Theo đề ta có:
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3\left(x-1\right)}\right)\)
ĐKXĐ : \(x\ne1;x\ne-4\)
\(45x-3\left(x-1\right)\left(x+4\right)=36\left(x-1\right)+12\left(x+4\right)\)
\(\Leftrightarrow45x+3x^2-3x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2-6x=0\)
\(\Leftrightarrow-3x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=-2\left(TMĐK\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
OY
29 tháng 1 2018
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow\dfrac{3(12x-x^2+4)}{3\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{4x+1}{3\left(x+4\right)\left(x-1\right)}\right)\)
\(\Leftrightarrow-x^2+12x+4=16x+4\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-4\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
24 tháng 7 2021
e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)
\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)
\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)
\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)