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1.A sai đề ?
1.B : \(x^2+x+6+2x\sqrt{x+3}=4\left(x+\sqrt{x+3}\right)\)
\(\Leftrightarrow x^2+x+6+2x\sqrt{x+3}=4x+4\sqrt{x+3}\)
\(\Leftrightarrow x^2+x+6+2x\sqrt{x+3}-4x-4\sqrt{x+3}=0\)
\(\Leftrightarrow x^2-3x+6+2x\sqrt{x+3}-4\sqrt{x+3}=0\)
\(\Leftrightarrow x^2-3x+6+2\sqrt{x+3}\left(x-2\right)=0\)
\(\Leftrightarrow x+3+2\sqrt{x+3}\left(x-2\right)+\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(\sqrt{x+3}+x-2\right)^2-1=0\)
\(\Leftrightarrow\left(\sqrt{x-3}+x-3\right)\left(\sqrt{x-3}+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}+x-3=0\\\sqrt{x-3}+x-1=0\end{matrix}\right.\)
Đến đây dễ rồi
Đáp án : \(\left[{}\begin{matrix}x=3\\x=\varnothing\end{matrix}\right.\)
2.A đang nghĩ
2.B
Áp dụng bất đẳng thức Cô-si :
\(\frac{x}{\sqrt{4x-1}}+\frac{\sqrt{4x-1}}{x}\ge2\sqrt{\frac{x\left(\sqrt{4x-1}\right)}{\left(\sqrt{4x-1}x\right)}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{\sqrt{4x-1}}=\frac{\sqrt{4x-1}}{x}\)
\(\Leftrightarrow x^2=4x-1\)
\(\Leftrightarrow x^2-4x+1=0\)
\(\Leftrightarrow x=2\pm\sqrt{3}\)( thỏa )
Vậy....
Đặt \(t=\sqrt{x}+\sqrt{1-x}\)\(\Rightarrow t^2=1+2\sqrt{x\left(1-x\right)}\)(\(t\ge0\))
\(pt:1+\frac{2}{3}\sqrt{x\left(1-x\right)}=\sqrt{x}+\sqrt{1-x}\)(\(0\le x\le1\))
\(\Leftrightarrow\frac{1}{3}\left(1+2\sqrt{x\left(1-x\right)}\right)+\frac{2}{3}=\sqrt{x}+\sqrt{1-x}\)
\(\Leftrightarrow\frac{1}{3}t^2+\frac{2}{3}=t\)
\(\Leftrightarrow t^2+2-3t=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1=\sqrt{x}+\sqrt{1-x}\\2=\sqrt{x}+\sqrt{1-x}\end{matrix}\right.\)
TH1:\(1=\sqrt{x}+\sqrt{1-x}\Leftrightarrow1=1+\sqrt{x\left(1-x\right)}\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
TH2:\(2=\sqrt{x}+\sqrt{1-x}\Leftrightarrow4=1+\sqrt{x\left(1-x\right)}\Leftrightarrow3=\sqrt{x\left(1-x\right)}\)
\(-x^2+x-9=0\)(vô nghiệm)
Vậy pt có nghiệm x = 0 , x = 1 .
ĐKXĐ: \(x\ge-\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}x+1=a>0\\\sqrt{2x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2+b^2}\)
\(\Leftrightarrow a^2+2ab+b^2=3a^2+b^2\)
\(\Leftrightarrow a^2-ab=0\Rightarrow\left[{}\begin{matrix}a=0\left(l\right)\\a=b\end{matrix}\right.\)
\(\Leftrightarrow x+1=\sqrt{2x+1}\)
\(\Leftrightarrow x^2+2x+1=2x+1\)
\(\Leftrightarrow x=0\)
