Rút gọn B = x +3/3 . (√) 18/50(x² + 6x +9) với x
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9(x + 1)² - 16(y + 3)²
= 9(x² + 2x + 1) - 16(y² + 6y + 9)
= 9x² + 18x + 9 - 16y² - 96y - 144
= 9x² - 16y² + 18x - 96y - 135
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)
\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)
\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)
\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)
\(\Leftrightarrow M=\frac{x-9}{2x}\)
Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)
\(B=\left(2x-3\right)\left(4x^2+6x+9\right)-\left(x+1\right)^2-\left(x-2\right)^3\)
\(=8x^3-27-x^2-2x-1-x^3+6x^2-12x+8\)
\(=7x^3+5x^2-14x-20\)
\(x-3-\sqrt{x^2-6x+9}\left(1\right)=x-3-\sqrt{\left(x-3\right)^2}=x-3-\left|x-3\right|\)
TH1: \(x< 3\)
\(\left(1\right)=x-3+x-3=2x-6\)
TH2: \(x\ge3\)
\(\left(1\right)=x-3-x+3=0\)
\(x-3-\sqrt{x^2-6x+9}\)
\(=x-3-\left|x-3\right|\)
\(=\left[{}\begin{matrix}x-3-x+3=0\left(x\ge3\right)\\x-3+x-3=2x-6\left(x< 3\right)\end{matrix}\right.\)
cho − 3 bé hơn bằng x bé hơn bằng 3 rút gọn biểu thức T= \(\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\) ta được
Ta có: \(T=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)
\(=\left|x-3\right|+\left|x+3\right|\)
\(=3-x+x+3\)
\(=6\)
\(B=\dfrac{x+3}{3}\cdot\sqrt{\dfrac{18}{50\left(x^2+6x+9\right)}}\)
\(=\dfrac{x+3}{3}\cdot\dfrac{3}{5\left(x+3\right)}=\dfrac{1}{5}\)