Tháng Tư 3, 2026

Thẻ: Toán lớp 8

Hỏi Đáp

Thực hiện phép tính: a) \(\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\,\,\,:\,\,\frac{{10x – 10y}}{{{x^3} + {y^3}}}\) b) \(\frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3{x^2} – 3x + 3}}{{{x^2} – 36}} + \frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3x}}{{{x^2} – 36}}\) c) \(\frac{{x – 1}}{{{x^2} – 4x + 4}}\, \cdot \,\frac{{{x^2} – 4}}{{{x^3} – 1}}\, \cdot \,\frac{{{x^2} + x + 1}}{{x + 2}}\) d) \(\left( {\frac{{3x}}{{1 – 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 – 6x + 9{x^2}}}\)

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Thực hiện phép tính: a) \(\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\,\,\,:\,\,\frac{{10x – 10y}}{{{x^3} + {y^3}}}\) b) \(\frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot …

Hỏi Đáp

Rút gọn biểu thức: a) \(\left( {\frac{9}{{{x^3} – 9x}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{{x^2} + 3x}} – \frac{x}{{3x + 9}}} \right)\) b) \(\frac{{x – 2}}{{{x^2} + 3x + 2}}\, \cdot \,\frac{{x + 2}}{{{x^2} – 5x + 6}}\) c) \(\frac{{{x^2} + 1}}{{3x}}:\frac{{{x^2} + 1}}{{x – 1}}:\frac{{{x^3} – 1}}{{{x^2} + x}}:\frac{{{x^2} + 2x + 1}}{{{x^2} + x + 1}}\) d) \(\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}}\)

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Rút gọn biểu thức: a) \(\left( {\frac{9}{{{x^3} – 9x}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{{x^2} + 3x}} – \frac{x}{{3x + 9}}} \right)\) b) …

Hỏi Đáp

Tìm \(P\) biết: \(\begin{align}& a)\frac{x-1}{{{x}^{2}}-x+1}-P=\frac{2}{x-1}+\frac{3x}{1-{{x}^{3}}} \\ & b)P+\frac{4x-12}{{{x}^{3}}-3{{x}^{2}}-4x+12}=\frac{3}{x-3}-\frac{{{x}^{2}}}{4-{{x}^{2}}} \\ \end{align}\)

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Tìm \(P\) biết: \(\begin{align}& a)\frac{x-1}{{{x}^{2}}-x+1}-P=\frac{2}{x-1}+\frac{3x}{1-{{x}^{3}}} \\ & b)P+\frac{4x-12}{{{x}^{3}}-3{{x}^{2}}-4x+12}=\frac{3}{x-3}-\frac{{{x}^{2}}}{4-{{x}^{2}}} \\ \end{align}\) A. \(\begin{array}{l} a)\,\,\,P = – \frac{1}{{x – 1}}\\ b)\,\,P = \frac{x}{{x + 3}} \end{array}\) B. …