Question

[tràn thị trúc oanh]

giải phương trình \(\left(x^2+3x+\frac{1}{4}\right)\left(x^2-x+\frac{1}{4}\right)=12x^2\)

tính giá trị biểu thức B=\(\frac{2x^2-5x\sqrt{y}+3y}{x\sqrt{y}-y}\) biết x=\(\sqrt{3+\sqrt{13+\sqrt{48}}}\) và y=4-2\(\sqrt{3}\)

Answer 1

[Nguyễn Việt Lâm]

Câu 1:

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{\left(x^2+\frac{1}{4}+3x\right)}{x}.\frac{\left(x^2+\frac{1}{4}-x\right)}{x}=12\)

\(\Leftrightarrow\left(x+\frac{1}{4x}+3\right)\left(x+\frac{1}{4x}-1\right)-12=0\)

Đặt \(x+\frac{1}{4x}-1=a\) ta được:

\(\left(a+4\right)a-12=0\Leftrightarrow a^2+4a-12=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4x}-1=2\\x+\frac{1}{4x}-1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+\frac{1}{4}=0\\x^2+5x+\frac{1}{4}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}...\\...\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

Câu 2:

\(x=\sqrt{3+\sqrt{12+2\sqrt{12}+1}}=\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)

\(=\sqrt{4+\sqrt{12}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

\(y=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{y}=\sqrt{3}-1\)

\(B=\frac{2\left(4+2\sqrt{3}\right)-5\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)+3\left(4-2\sqrt{3}\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\left(4-2\sqrt{3}\right)}\)

\(B=\frac{8+4\sqrt{3}-10+12-6\sqrt{3}}{2-4+2\sqrt{3}}=\frac{10-2\sqrt{3}}{-2+2\sqrt{3}}=\frac{5-\sqrt{3}}{\sqrt{3}-1}\)

\(B=\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{5\sqrt{3}+5-3-\sqrt{3}}{2}=\frac{2+4\sqrt{3}}{2}=2\sqrt{3}+1\)