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\(\frac{x^3+125}{x^2-3x-40}=\frac{x^3+5^3}{\left(x^2+5x\right)-\left(8x+40\right)}=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{x\left(x+5\right)-8\left(x+5\right)}\)
\(=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{\left(x+5\right)\left(x-8\right)}=\frac{x^2-5x+25}{x-8}\)
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phân tích thành nhân tử ở mẫu và tử sau đó ta rút gọn vậy là ra đáp số
a) \(=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(4x-5\right)}\)\(\)
\(=\frac{5x\cdot\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(=\frac{5x\left(4x+5\right)}{x-3}\)
b) \(=\frac{3^2-\left(x+5\right)^2}{\left(x+2\right)^2}\)
\(=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)
\(=\frac{\left(x+2\right)\left(8+x\right)}{\left(x+2\right)^2}\)
\(=\frac{8+x}{x+2}\)
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\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)
\(=\frac{x\left(x+1\right)+\left(x+1\right)}{x\left(x-1\right)+2x^2-2x+x+1}\)
\(=\frac{\left(x+1\right)\left(x+1\right)}{x\left(x-1\right)+2\left(x-1\right)+\left(x+1\right)}\)
Ddeeff sao rồi bạn ko rút gọn được
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\(=\frac{x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(=\frac{x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\frac{x\left(4x+5\right)}{x-3}=\)\(\frac{4x^2-5x}{x-3}\)
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\(\frac{x^4-y^4}{y^3-x^3}\)
\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}\)
\(=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(y^2+xy+x^2\right)}\)
\(=-\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{x^2+xy+y^2}\)