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a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
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Bạn sửa lại đề dùm mình nha, sai đề hơi nhiều đó.
ĐKXĐ:\(x\ne0;2\)
\(P=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2+2x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}-\frac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right).\frac{x^2-x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}+\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right).\frac{x^2-2x+x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)^2}{2\left(x^2+4\right)\left(x-2\right)}+\frac{4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right).\frac{x\left(x-2\right)+\left(x-2\right)}{x^2}\)
\(P=\frac{x\left(x^2-4x+4\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{x^3-4x^2+4x-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{\left(x^3+4x\right)\left(x-2\right)\left(x+1\right)}{2\left(x^2+4\right)\left(x-2\right).x^2}\\ P=\frac{x\left(x^2+4\right)\left(x-2\right)\left(x+1\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\\ P=\frac{x+1}{2x}\)
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A) x2 - 4x + 4 = (x - 2)2 (hằng đẳng thức số 2)
Cm : x2 - 4x + 4 = x2 - 2x - 2x + 4 = x(x - 2) - 2(x - 2) = (x - 2)(x - 2) = (x - 2)2
b tương tự
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a) Ta có: \(x\left(x+3\right)-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
b) Ta có: \(4x^2-1+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x\right)^2-1^2+x\left(2x-1\right)=0\)
\(\Rightarrow\left[\left(2x\right)^2-1^2\right]+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)+x\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{-1}{3}\right\}\)
a, x(x+3)-2x-6=0
⇔x(x+3)-2(x+3)=0
⇔(x+3)(x-2)=0
⇔x+3=0 hoặc x-2=0
⇔x= -3 hoặc x=2
b, 4x2-1+x(2x-1)=0
⇔(2x-1)(2x+1)+x(2x-1)=0
⇔(2x-1)(2x+1+x)=0
⇔(2x-1)(3x+1)=0
⇔2x-1=0 hoặc 3x+1=0
⇔x=1/2 hoặc x= -1/3
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\(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)+4\left(3-x\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}}\)
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
\(=x^2-2x+4\)
= x2 - 2x + 4