a/ \(-4\le x\le1\)
\(\Leftrightarrow1-x+4+x+2\sqrt{4-3x-x^2}=9\)
\(\Leftrightarrow\sqrt{4-3x-x^2}=2\)
\(\Leftrightarrow-3x-x^2=0\)
\(\Leftrightarrow x\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ge-\frac{3}{2}\)
\(\Leftrightarrow x^2+2x+1+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
c/ ĐKXĐ: \(x;y\ge4\)
\(\Leftrightarrow\frac{2\left(x\sqrt{y-4}+y\sqrt{x-4}\right)}{xy}=1\)
\(\Leftrightarrow\frac{2\sqrt{y-4}}{y}+\frac{2\sqrt{x-4}}{x}=1\)
Mặt khác theo BĐT Cauchy:
\(\frac{2\sqrt{y-4}}{y}+\frac{2\sqrt{x-4}}{x}\le\frac{2^2+y-4}{2y}+\frac{2^2+x-4}{2x}=\frac{1}{2}+\frac{1}{2}=1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{y-4}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=8\end{matrix}\right.\)
a,\(\sqrt{1-x}-1+\sqrt{4+x}-2=0\)\(\Leftrightarrow x\left(\frac{1}{\sqrt{1-x}+1}+\frac{1}{\sqrt{4+x}+2}\right)=0\Rightarrow x=0\)
b,\(\Leftrightarrow x^2+6x+9=2x+3+2\sqrt{2x+3}+1\)
\(\Leftrightarrow\left(x+3\right)^2=\left(\sqrt{2x+3}+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3-\sqrt{2x+3}-1=0\\x+3+\sqrt{2x+3}+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2x+3}+2=0\\x+\sqrt{2x+3}+4=0\end{matrix}\right.\)
a) \(\sqrt{1-x}+\sqrt{4-x}=3\)
ĐKXĐ:\(x\ge1\)
\(\Leftrightarrow\sqrt{1-x}=3-\sqrt{4-x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\\left(\sqrt{1-x}\right)^2=\left(9-6\sqrt{x-4}+4-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\1-x=13-x-6\sqrt{x-4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\6\sqrt{x-4}=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\sqrt{x-4}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=8\left(TM\right)\end{matrix}\right.\)
Vậy x=8
b)\(x^2+4x+5=2\sqrt{2x+3}\)
ĐKXĐ:\(x\ge\frac{-3}{2}\)
\(\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\Leftrightarrow x=-1\left(TM\right)\)
Vậy x=-1
a) 1−x+4−x=3
ĐKXĐ:x≥1
⇔1−x=3−4−x
⇔{1−x≥0(1−x)2=(9−6x−4+4−x)
⇔{x≥11−x=13−x−6x−4
⇔{x≥16x−4=12
⇔{x≥1x−4=2
⇔{x≥1x=8(TM)
Vậy \(x=8 \)
b)x2+4x+5=22x+3
ĐKXĐ:x≥−32
⇔x2+4x+5−22x+3
⇔(x2+2x+1)+(2x+3−22x+3+1)=0
⇔(x+1)2+(2x+3−1)2=0
⇔[x+1=02x+3−1=0⇔x=−1(TM)
Vậy \( x=-1 \)
