Question

\(\sqrt{64x-64}-25\sqrt{x+1}+\sqrt{4x+4}=20\)

Tìm đk và tìm x

Answer 1

`x >=-1`

`8 sqrt(x+1) - 25sqrt(x+1) + 2 sqrt(x+1) = 20`

`-> -15 sqrt(x+1) = 20`

`-> sqrt(x+1) = -4/3`

`-> x = cancel o`

Answer 2

\(\sqrt{64x-64}-25\sqrt{x+1}+\sqrt{4x+4}=20\) (\(ĐK:x\ge1\)

\(8\sqrt{x-1}-23\sqrt{x+1}=20\)

Vì \(8\sqrt{x-1}+23\sqrt{x+1}\ne0\)

pt\(\Leftrightarrow\) \(\dfrac{64\left(x-1\right)-529\left(x+1\right)}{\text{ }8\sqrt{x-1}+23\sqrt{x+1}}=20\)

    \(\dfrac{-465x-593}{8\sqrt{x-1}+23\sqrt{x+1}}=20\)

Vì \(x\ge1\)

\(\Rightarrow VT< 0\Rightarrow VT\ne VP\)

\(\Rightarrow\) phương trình vô nghiệm

Answer 3

đk x ≥ -1

Sửa đề 

\(\sqrt{64x+64}-25\sqrt{x+1}+\sqrt{4x+4}=20\\ 8\sqrt{x+1}-25\sqrt{x+1}+2\sqrt{x+1}=20\\ -15\sqrt{x+1}=20\\ \sqrt{x+1}=-\dfrac{20}{15}\left(voli\right)\)