Question

tìm x

\(5\sqrt{x-2}=10+\sqrt{9x-18}\)

Answer 1

Ta có:5\(\sqrt{x-2}\)=10+\(\sqrt{9\text{x}-18}\)

<=>5\(\sqrt{x-2}\)=10+\(\sqrt{9\left(x-2\right)}\)

<=>5\(\sqrt{x-2}\)=10+3\(\sqrt{x-2}\)

<=>5\(\sqrt{x-2}\) - 3\(\sqrt{x-2}\)=10

<=>2\(\sqrt{x-2}\)=10

<=>\(\sqrt{x-2}\)=5

<=>\(\sqrt{x-2}\)=\(\sqrt{25}\)

<=>x-2=25

<=>x=25+2=27

Vậy x có giá trị là 27

Answer 2

Tìm \(x\)

\(5\sqrt{x-2} = 10+\sqrt{9x-18}\)

<=> \(5\sqrt{x-2} = 10+\sqrt{9(x-2)}\)

<=> \(5\sqrt{x-2} = 10+3\sqrt{x-2}\)

<=> \(5\sqrt{x-2} - 3\sqrt{x-2} = 10\)

<=> \(2\sqrt{x-2} = 10\)

<=> \(\sqrt{x-2} = 5\)

<=> \(\sqrt{(x-2)^2} = 5^2\)

<=> \(|x-2|=25\)

* \(x-2=25\)

<=> \(x=27\)

* \(-x-2=25\)

<=> \(-x=27\)

<=> \(x=-27\)

Vậy \(x = 27 \) hoặc \(x=-27\)