Question

Bài 1: Tìm x,y:

a) |x - 1| + |x + 3| = 4

b) |2x + 3| + |2x - 1| = \(\frac{8}{2\left(y-5\right)^2+2}\)

c) |x + 3| + |x + 1| = \(\frac{16}{\left|y-2\right|+\left|y+2\right|}\)

Bài 2: Tìm số nguyên x,y, biết: 

a) \(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\)

b) \(x^2-2xy+y=0\)

Answer 1

a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)

\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)

b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)

c Tương tự b

2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)

\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)

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