Question

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

làm nhanh hộ mình với cần rất chi là gấp

Answer 1

Dễ mà.

Đk: \(x\ge-5\)

   \(\sqrt{x+5}=1-x\)

\(\Leftrightarrow x+5=1-2x+x^2\)

\(\Leftrightarrow x^2-3x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\left(tm\right)\\x=-1\left(tm\right)\end{cases}}\)

       Vậy x = {-1;4}

Answer 2

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

ĐK :\(\hept{\begin{cases}x+5\ge0\Rightarrow x\ge-5\\\sqrt{x+5}\ge0\Rightarrow1-x\ge0\Rightarrow x\le1\end{cases}\Rightarrow-5\le x\le1}\)

=> x + 5 = (1 - x)² = 1 - 2x + x²

=> 5     = 1 - 3x + x²

=> 6,25 = 2,25 - 3x + x²

6,25 = (1,5)² - 2.1,5.x + x²

6,25 = (1,5 - x)

\(\Rightarrow\orbr{\begin{cases}1,5-x=-2,5\\1,5-x=2,5\end{cases}\Rightarrow\orbr{\begin{cases}x=4\left(ktmđk\right)\\x=-1\left(tmđk\right)\end{cases}}}\).Vậy x = -1

Answer 3

\(\Leftrightarrow x+5=1-2x+x^2\)

\(\Leftrightarrow-x^2+3x+4=0\)

\(\Leftrightarrow-x^2-x+4x+4=0\)

\(\Leftrightarrow-x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\4-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\x=4\end{cases}}}\)

vậy..........

Answer 4

\(\Leftrightarrow x+5=1-2x+x^2\)

\(\Leftrightarrow-x^2+3x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\4-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\x=4\end{cases}}}\)

vậy..........