Question

Tìm x, biết:

b)\(5\sqrt{x-1}-\sqrt{36x-36}\)\(+\sqrt{9x-9}=\sqrt{8x+12}\)

Answer 1

Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\)   \(\left(ĐK:x\ge1\right)\)

    \(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)

    \(\Leftrightarrow4.\left(x-1\right)=8x+12\)

    \(\Leftrightarrow4x-4=8x+12\)

    \(\Leftrightarrow-4x=16\)

    \(\Leftrightarrow x=-4\left(L\right)\)

Vậy \(S=\varnothing\)

Answer 2

\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\) 

\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\) 

\(2\sqrt{x-1}=2\sqrt{2x+3}\) 

\(\sqrt{x-1}=\sqrt{2x+3}\) 

\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\) 

\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\) 

\(\Rightarrow x=2\)

Answer 3

Bạn FL.capheny bị sai dấu 2x+3 thành 2x-3 mà ai vẫn tk đúng vậy

Answer 4

Á nhầm sửa lại 

\(\hept{\begin{cases}2x+3\ge0\\x-1=2x+3\end{cases}}\) 

\(\hept{\begin{cases}2x\ge-3\\-x=4\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=-4\end{cases}}\) 

\(\Rightarrow x=\varnothing\)