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a) ĐKXĐ: \(x\notin\left\{0;2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
Suy ra: \(x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-1}
a) \(⇔x^2-9x+20=12 \)
\(⇔x^2-9x+8=0\)
\(⇔x^2-x-8x+8=0\)
\(⇔(x-1)(x-8)=0\)
\(⇔\left[\begin{array}{} x-1=0\\ x-8=0 \end{array}\right.⇔\left[\begin{array}{} x=1\\ x=8 \end{array}\right.\)
b) \(⇔4x^2-12x+8=3\)
\(⇔4x^2-12x+5=0\)
\(⇔(2x-1)(2x-5)=0\)
\(⇔\left[\begin{array}{} 2x-1=0\\ 2x-5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{5}{2} \end{array}\right.\)
c) \(⇔x^2+x-30=42\)
\(⇔x^2+x-72=0\)
\(⇔(x-9)(x+8)=0\)
\(⇔\left[\begin{array}{} x-9=0\\ x+8=0 \end{array}\right.⇔\left[\begin{array}{} x=9\\ x=-8 \end{array}\right.\)
d) \(⇔2x^2+5x-3=-6\)
\(⇔2x^2+5x+3=0\)
\(⇔(x+1)(2x+3)=0\)
\(⇔\left[\begin{array}{} x+1=0\\ 2x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=-\frac{3}{2} \end{array}\right.\)
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
1) \(x^2+x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x+3\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)
4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)
5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)
6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)
bài này tìm x hay tìm cực trị vậy