Question

Giải phương trình

a, \(2\sqrt{x-2}=16\)

b, \(\sqrt{x-1}>3\)

c, \(-5\sqrt{2x+4}\le-10\)

d, \(7-5\sqrt{2-x}=-8\)

e, \(\sqrt{x^2-14x+49}+x=7\)

Answer 1

\(a.2\sqrt{x-2}=16\left(ĐK:x\ge2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)

\(b.\sqrt{x-1}>3\left(ĐK:x\ge1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)

\(c.-5\sqrt{2x+4}\le-10\left(ĐK:x\ge2\right)\\ \Leftrightarrow\sqrt{2x+4}\ge2\\ \Leftrightarrow2x+4\ge4\\ \Leftrightarrow2x\ge0\Leftrightarrow x\ge0\)

Answer 2

\(a.2\sqrt{x-2}=16\left(ĐK:x>2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)

b.\(\sqrt{x-1}>3\left(ĐK:x>1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)

\(c.-5\sqrt{2x+4}< -10\left(ĐK:x>-2\right)\\ \Leftrightarrow\sqrt{2x+4}>2\\ \Leftrightarrow2x+4>4\\ \Leftrightarrow2x>0\Leftrightarrow x>0\)

Answer 3

\(d.7-5\sqrt{2-x}=-8\left(ĐK:x\le2\right)\\ \Leftrightarrow-5\sqrt{2-x}=-15\\ \Leftrightarrow\sqrt{2-x}=3\\ \Leftrightarrow2-x=9\Leftrightarrow x=-7\)

\(e.\sqrt{x^2-14x+49}+x=7\\ \Leftrightarrow\sqrt{x^2-2\cdot x\cdot7+\left(7\right)^2}+x=7\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}+x=7\\ \Leftrightarrow\left|x-7\right|+x-7=0\left(1\right)\)

Có 2 trường hợp:

+ T/h 1: \(x-7\ge0\Leftrightarrow x\ge7\)ta có:

\(\left(1\right)\Leftrightarrow x-7+x-7=0\Leftrightarrow2x-14=0\Leftrightarrow x=7\)

+ T/h 2 : \(x-7< 0\Leftrightarrow x< 7\)ta có:

\(\left(1\right)\Leftrightarrow7-x+x-7=0\Leftrightarrow0x=0\)(PT vô số nghiệm)

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