Question

Bài 47 (Đề thi thử 10 - THCS Nguyễn Tất Thành - Hà Nội 2017 - 2018) (2 điểm)

1. Rút gọn biểu thức \(P=\dfrac{1}{2\left(1+\sqrt{x}\right)}+\dfrac{1}{2\left(1-\sqrt{x}\right)}-\dfrac{x^2+2}{1-x^3}\)

2. Tính giá trị biểu thức \(A=a^3+3a+2003\) với \(a=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)

 

Answer 1

1: \(P=\dfrac{1-\sqrt{x}+1+\sqrt{x}}{2\left(1-x\right)}-\dfrac{x^2+2}{\left(1-x\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2}{2\left(1-x\right)}-\dfrac{x^2+2}{\left(1-x\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^2+x+1-x^2-2}{\left(1-x\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(1-x\right)\left(x^2+x+1\right)}=\dfrac{-1}{x^2+x+1}\)

Answer 2

`2)`\(a=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)

\(a^3=7+5\sqrt{2}+7-5\sqrt{2}+3\left(\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\right)\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)\(\Leftrightarrow a^3=14+3\sqrt[3]{49-50}a\)

\(\Leftrightarrow a^3=14-3a\)

\(\Leftrightarrow a^3+3a-14=0\)

\(A=a^3+3a+2003\)

\(A=\left(a^3+3a-14\right)+2017\)

\(A=0+2017\)

\(A=2017\)