Question

E=\(\frac{x^2}{x-2}\cdot\left(\frac{x^2+4}{x}-4\right)\) a)rút gọn b)tìm x để E=2 c)tính giá tri E biết \(\left|x+1\right|=2\) d)tìm giá trị nhỏ nhât của E

Answer 1

a)

\(E=\frac{x^2}{x-2}\cdot\left(\frac{x^2+4}{x}-4\right)\\ \Leftrightarrow\frac{x^2}{x-2}\cdot\left(x+\frac{4}{x}-4\right)\\ \Leftrightarrow\frac{x^2\cdot\left(x+\frac{4}{x}-4\right)}{x-2}=\frac{x^3+4x-4x^2}{x-2}\)

b)

\(E=\frac{x^3+4x-4x^2}{x-2}=2\\ \Leftrightarrow\frac{x\cdot\left(x^2+4x-4\right)}{x-2}=2\\ \)

\(\Leftrightarrow\frac{x\cdot\left(x-2\right)^2}{x-2}=2\\ \Leftrightarrow x\cdot\left(x-2\right)=2\\ \Leftrightarrow x^2-2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2+2\sqrt{3}}{2}\\x=\frac{2-2\sqrt{3}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)

Answer 2

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