GIÚPPPPPPPP
bài 1 tìm x bt :
a) √6-4x-x² =x+4
b) √x²+1-x=3
c) √x<4
d) √x-1>4
bài 2 tìm x để biểu thức sau có nghĩa :
a) √-3/x-5
b) √1-2x
c) 1/1- √1-1
d) 1/ √4-x
GIÚPPPPPPPP
bài 1 tìm x bt :
a) √6-4x-x² =x+4
b) √x²+1-x=3
c) √x<4
d) √x-1>4
bài 2 tìm x để biểu thức sau có nghĩa :
a) √-3/x-5
b) √1-2x
c) 1/1- √1-1
d) 1/ √4-x
Đáp án:
$\begin{array}{l}
B1)\\
a)\sqrt {6 – 4x – {x^2}} = x + 4\\
Dkxd:\left\{ \begin{array}{l}
6 – 4x – {x^2} \ge 0\\
x + 4 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 4x – 6 \le 0\\
x \ge – 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} \le 10\\
x \ge – 4
\end{array} \right.\\
\Rightarrow – 4 \le x \le \sqrt {10} – 2\\
Pt: \Rightarrow 6 – 4x – {x^2} = {\left( {x + 4} \right)^2}\\
\Rightarrow – {x^2} – 4x + 6 = {x^2} + 8x + 16\\
\Rightarrow 2{x^2} + 12x + 10 = 0\\
\Rightarrow {x^2} + 6x + 5 = 0\\
\Rightarrow \left( {x + 1} \right)\left( {x + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = – 1\left( {tmdk} \right)\\
x = – 5\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = – 1\\
b)\sqrt {{x^2} + 1 – x} = 3\\
Dkxd:{x^2} + 1 – x \ge 0\left( {\text{luôn}\,\text{đúng}} \right)\\
\Rightarrow {x^2} + 1 – x = 9\\
\Rightarrow {x^2} – x – 8 = 0\\
\Rightarrow {x^2} – 2.\dfrac{1}{2}.x + \dfrac{1}{4} = 8 + \dfrac{1}{4}\\
\Rightarrow {\left( {x – \dfrac{1}{2}} \right)^2} = \dfrac{{33}}{4}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {33} }}{4}\\
x = \dfrac{{1 – \sqrt {33} }}{4}
\end{array} \right.\\
c)\sqrt x < 4\left( {dk:x \ge 0} \right)\\
\Rightarrow x < 16\\
\text{Vậy}\,0 \le x < 16\\
d)\sqrt {x – 1} > 4\left( {dk:x \ge 1} \right)\\
\Rightarrow x – 1 > 16\\
\Rightarrow x > 17\\
\text{Vậy}\,x > 17\\
B2)\\
a)\sqrt {\dfrac{{ – 3}}{{x – 5}}} \\
\Rightarrow \dfrac{{ – 3}}{{x – 5}} \ge 0\\
\Rightarrow x – 5 < 0\\
\Rightarrow x < 5\\
b)1 – 2x \ge 0\\
\Rightarrow 2x \le 1\\
\Rightarrow x \le \dfrac{1}{2}\\
c)??\\
d)\dfrac{1}{{\sqrt {4 – x} }}\\
\Rightarrow 4 – x > 0\\
\Rightarrow x < 4
\end{array}$