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31 tháng 12 2021
a: \(P=\dfrac{x}{x+3}-\dfrac{x^2-5x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{3}{x-3}\)
\(=\dfrac{x^2-3x-x^2+5x+6+3x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{5}{x-3}\)
A
2
20 tháng 2 2022
\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{-x^2+2x+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
20 tháng 2 2022
\(\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\left(ĐKXĐ:x\ne2,x\ne3\right)\)
\(=\dfrac{2x-9}{x^2-3x-2x+6}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9}{x\left(x-3\right)-2\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{x-2}+\dfrac{\left(2x+1\right)\left(x-2\right)}{x-3}\)
\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{x-2}+\dfrac{2x^2-4x+x-2}{x-3}\)
\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x\left(x-2\right)+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x+1}{x-3}\)
LD
0
16 tháng 6 2018
a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)
TN
2 tháng 7 2017
\(1,\left(2-x\right)^2-9=0\)
\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)
\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)
2 tháng 7 2017
1. \(\left(2-x\right)^2-9=0\)
\(\left(2-x\right)^2=9\)
\(\left(2-x\right)^2=3^2\)
\(2-x=3\)
\(-x=-1\Rightarrow x=1\)
17 tháng 3 2020
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(x^3-2x^2+4x+2x^2-4x+8-x^3+2x=15\)
\(2x+8=15\)
\(2x=7\)
\(x=\frac{7}{2}\)
17 tháng 3 2020
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\frac{10}{9}\)
6 tháng 1 2022
a: Thay x=-4 vào B, ta được:
\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)
b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)
\(=\dfrac{x^2+2x}{\left(x-3\right)}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)
c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
ML
1
28 tháng 5 2021
Đề như này hả bạn? :V
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow-9x^2+27x+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow45x=6\)\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy...
