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ĐKXĐ: ...
\(\Leftrightarrow5x+9=x+5+4\sqrt{x+1}\)
\(\Leftrightarrow x+1=\sqrt{x+1}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
1.
a/ ĐKXĐ: \(-1\le x\le5\)
\(\Leftrightarrow\sqrt{x+3}\le\sqrt{5-x}+\sqrt{x+1}\)
\(\Leftrightarrow x+3\le6+2\sqrt{\left(5-x\right)\left(x+1\right)}\)
\(\Leftrightarrow x-3\le2\sqrt{-x^2+4x+5}\)
- Với \(x< 3\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP\ge0\end{matrix}\right.\) BPT luôn đúng
- Với \(x\ge3\) cả 2 vế ko âm, bình phương:
\(x^2-6x+9\le-4x^2+16x+20\)
\(\Leftrightarrow5x^2-22x-11\le0\) \(\Rightarrow\frac{11-4\sqrt{11}}{5}\le x\le\frac{11+4\sqrt{11}}{5}\)
\(\Rightarrow3\le x\le\frac{11+4\sqrt{11}}{5}\)
Vậy nghiệm của BPT đã cho là \(-1\le x\le\frac{11+4\sqrt{11}}{5}\)
1b/
Đặt \(\sqrt{2x^2+8x+12}=t\ge2\)
\(\Rightarrow x^2+4x=\frac{t^2}{2}-6\)
BPT trở thành:
\(\frac{t^2}{2}-12\ge t\Leftrightarrow t^2-2t-24\ge0\) \(\Rightarrow\left[{}\begin{matrix}t\le-4\left(l\right)\\t\ge6\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+8x+12}\ge6\)
\(\Leftrightarrow2x^2+8x-24\ge0\Rightarrow\left[{}\begin{matrix}x\le-6\\x\ge2\end{matrix}\right.\)
\(y=\frac{\sqrt{2017\left(x-2015\right)}}{\sqrt{2017}\left(x+2\right)}+\frac{\sqrt{2016\left(x-2016\right)}}{\sqrt{2016}x}\le\frac{1}{2\sqrt{2017}}+\frac{1}{2\sqrt{2016}}\)
"=" \(\Leftrightarrow\)\(x=4032\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\sqrt{3+x}+\sqrt{6-x}=3\)(ĐK: \(-3\le x\le6\))
\(\Leftrightarrow\left(\sqrt{3+x}+\sqrt{6-x}\right)^2=3^2\)
\(\Leftrightarrow3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(3+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=6\end{cases}}\)(thỏa mãn)