Cho biểu thức B = (\(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\)) : \(\frac{6x-12}{2x^2-18}\)
a, Tìm TẬP XÁC ĐỊNH và RÚT GỌN biểu thức B
b, Tìm GIÁ TRỊ của B với \(|x=1|=2\)
c, Tìm GIÁ TRỊ NGUYÊN của X để B nhận GIÁ TRỊ NGUYÊN
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a) ĐKXĐ: x - 3 \(\ne\)0 x \(\ne\)3
9 - x2 \(\ne\)0 <=> x \(\ne\)\(\pm\)3
x + 3 \(\ne\)0 x \(\ne\)-3
\(\frac{6x-12}{2x^2-18}\) \(\ne\)0 \(6x-12\ne0\) và \(2x^2-18\ne0\)
x \(\ne\)\(\pm\)3
<=> \(x\ne2\) và x \(\ne\)\(\pm\)3
<=> x \(\ne\)\(\pm\)3 và x \(\ne\)2
Ta có: B = \(\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)
B = \(\left(\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{6\left(x-2\right)}{2\left(x^2-9\right)}\)
B = \(\left(\frac{x^2+6x+9-2x^2+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{3x+15}{\left(x+3\right)\left(x-3\right)}\cdot\frac{\left(x-3\right)\left(x+3\right)}{3\left(x-2\right)}\)
B = \(\frac{3\left(x+5\right)}{3\left(x-2\right)}\)
B = \(\frac{x+5}{x-2}\)
b) (sai đề)
c) Ta có: B = \(\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để B \(\in\)Z <=> 7 \(⋮\)x - 2 <=> x - 2 \(\in\)Ư(7) = {1; -1; 7; -7}
Lập bảng:
Vậy ...
a) \(\text{ĐKXĐ:}\hept{\begin{cases}x\ne\pm3\\x\ne2\end{cases}}\)
\(B=\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)
\(B=\left[\frac{x+3}{x-3}+\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\left[\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]\)
\(B=\left[\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\frac{x^2+6x+9-\left(2x^2-6\right)+x^2-3}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)
\(B=\frac{3\left(x+5\right)}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x-3\right)\left(x+3\right)}{6\left(x-2\right)}\)
\(B=\frac{x+5}{x-2}\)
b) Ta có: \(\frac{x+5}{x-2}=1+\frac{7}{x-2}\)
Để B nguyên thì: \(7⋮x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)\)
\(\RightarrowƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng:
Vậy: \(x\in\left\{1;-5;9\right\}\)