Ai giúp e với ạ e cảm ơn làm câu a là được rồi ạ
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12 tháng 11 2021
Ta có: 1(x−2)2−1x+2−14−x21(x−2)2−1x+2−14−x2
=x+2−(x−2)2+x−2(x−2)2⋅(x+2)=x+2−(x−2)2+x−2(x−2)2⋅(x+2)
=2x−x2+4x−4(x−2)2⋅(x+2)=2x−x2+4x−4(x−2)2⋅(x+2)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
12 tháng 11 2021
a) (3x³ - 5x² + 9x - 15) : (3x - 5)
= [x²(3x - 5) + 3(3x - 5)] : (3x - 5)
= [(3x - 5)(x² + 3)] : (3x - 5)
= x² + 3
CT
12 tháng 11 2021
a) (3x³ - 5x² + 9x - 15) : (3x - 5)
= [x²(3x - 5) + 3(3x - 5)] : (3x - 5)
= [(3x - 5)(x² + 3)] : (3x - 5)
= x² + 3
LH
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{2}{x-3}+\frac{x-3}{x+3}-\frac{x-15}{9-x^2}\left(x\ne\pm3\right)\)
\(=\frac{2}{x-3}+\frac{x-3}{x+3}-\frac{x}{\left(3-x\right).\left(3+x\right)}\)
\(=\frac{2}{x-3}+\frac{x-3}{x+3}+\frac{x-15}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{2.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}+\frac{\left(x-3\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}+\frac{x-15}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{2x+6+\left(x-3\right)^2+x-15}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{2x+6+x^2-6x+9+x-15}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{x^2+\left(2x-6x+x\right)+\left(6+9-15\right)}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{x^2-3x}{\left(x-3\right).\left(x+3\right)}\)
\(=\frac{x}{x+3}\)
b) Thay vào ta được:
\(\frac{1}{2}:\left(\frac{1}{2}+3\right)=\frac{1}{7}\)
Vậy biểu thức \(A=\frac{1}{7}\) khi \(x=\frac{1}{2}\)