Question

1.Giải phương trình \(x^2-7x=6\sqrt{x+5}-30\)

2. cho x, y là các số dương thỏa mãn \(x+y\ge\frac{34}{35}\) tìm min của \(A=3x+4y+\frac{2}{5x}+\frac{8}{7y}\)

Answer 1

a/ ĐKXĐ: ....

\(\Leftrightarrow x^2-8x+16+x+14-6\sqrt{x+5}=0\)

\(\Leftrightarrow\left(x-4\right)^2+\frac{\left(x+14\right)^2-36\left(x+5\right)}{x+14+6\sqrt{x+5}}=0\)

\(\Leftrightarrow\left(x-4\right)^2+\frac{x^2-8x+16}{x+14+6\sqrt{x+5}}=0\)

\(\Leftrightarrow\left(x-4\right)^2\left(1+\frac{1}{x+14+6\sqrt{x+5}}\right)=0\)

2/

\(A=\frac{5x}{2}+\frac{2}{5x}+\frac{7y}{2}+\frac{8}{7y}+\frac{1}{2}\left(x+y\right)\)

\(A\ge2\sqrt{\frac{10x}{10x}}+2\sqrt{\frac{56y}{14y}}+\frac{1}{2}.\frac{34}{35}=\frac{227}{35}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{2}{5}\\y=\frac{4}{7}\end{matrix}\right.\)

Answer 2

1.

\(PT\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{x+5}-3\right)^2=0\left(x\ge-5\right)\)

\(\Leftrightarrow x-4=\sqrt{x+5}-3=0\Leftrightarrow x=4\).