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\(\sqrt{\frac{1}{4}x^2+x+1}=\sqrt{\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.1+1^2}=\sqrt{\left(\frac{1}{2}x+1\right)^2}=\left|\frac{1}{2}x+1\right|\)
\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
phương trình <=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)
<=> \(\frac{1}{2}x+1=\sqrt{5}-1\) hoặc \(\frac{1}{2}x+1=-\sqrt{5}+1\)
+) \(\frac{1}{2}x+1=\sqrt{5}-1\)<=> \(x=2\sqrt{5}+4\)
+) \(\frac{1}{2}x+1=-\sqrt{5}+1\) <=> \(x=-2\sqrt{5}\)
Vậy pt có 2 nghiệm \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)
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a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)
\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=4\)
\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=4\)
Làm nốt
b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)
Làm nốt
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\(\Leftrightarrow\left(x-\frac{1}{x}\right)+\sqrt{x-\frac{1}{x}}=\left(2x-\frac{5}{x}\right)+\sqrt{2x-\frac{5}{x}}\)
\(a=\sqrt{x-\frac{1}{x}};\text{ }b=\sqrt{2x-\frac{5}{2}};\text{ }a,\text{ }b>0\)
\(a^2+a=b^2+b\Leftrightarrow\left(a-b\right)\left(a+b\right)+\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\text{ }\left(do\text{ }a+b+1\ge1>0\right)\)
\(\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\Leftrightarrow x-\frac{4}{x}=0\Leftrightarrow x^2-4=0\Leftrightarrow x=\pm2\)
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2/ x2 - 6x + 4 + \(2\sqrt{2x-1}\)= 0
<=> (x2 - 4x + 4) - (2x - 1 - \(2\sqrt{2x-1}\)+1) = 0
<=> (x - 2)2 - (1 - \(\sqrt{2x-1}\))2 = 0
\(\Leftrightarrow\left(x-1-\sqrt{2x-1}\right)\left(x-3+\sqrt{2x-1}\right)=0\)
Làm tiếp nhé
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ĐKXĐ:\(\hept{\begin{cases}x-2>0\\y-1>0\\z-5>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>2\\y>1\\z>5\end{cases}}\)
pt\(\Leftrightarrow\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}+\sqrt{x-2}+\sqrt{y-1}+\sqrt{z-5}=16\)
Áp dụng BĐT Cauchy:
\(\frac{4}{\sqrt{x-2}}+\sqrt{x-2}+\frac{1}{\sqrt{y-1}}+\sqrt{y-1}+\frac{25}{\sqrt{z-5}}+\sqrt{z-5}\)
\(\ge2\sqrt{\frac{4}{\sqrt{x-2}}.\sqrt{x-2}}+2\sqrt{\frac{1}{\sqrt{y-1}}.\sqrt{y-1}}+2\sqrt{\frac{25}{\sqrt{z-5}}.\sqrt{z-5}}\)
\(=2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2.2+2.1+2.5\)
\(=4+2+10=16\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=4\\y-1=1\\z-5=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\\z=30\end{cases}}\)
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Cần gì phải thế.
Đặt \(\sqrt{x-\frac{1}{x}}=a\ge0;\sqrt{2x-\frac{5}{x}}=b\ge0\Rightarrow x-\frac{4}{x}=b^2-a^2\)
\(\Rightarrow a=b^2-a^2+b\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
Đến đây tự làm tiếp
ĐKXĐ: \(x>0\)
\(\Leftrightarrow x+\frac{1}{x}-4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+5=0\)
Đặt \(\sqrt{x}+\frac{1}{\sqrt{x}}=t>0\Rightarrow t^2=x+\frac{1}{x}+2\Rightarrow x+\frac{1}{x}=t^2-2\)
Pt trở thành:
\(t^2-2-4t+5=0\Leftrightarrow t^2-4t+3=0\) \(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{\sqrt{x}}=1\\\sqrt{x}+\frac{1}{\sqrt{x}}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}+1=0\left(vn\right)\\x-3\sqrt{x}+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)