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ĐKXĐ: \(x^2-4x+1\ge0\)
\(2x+2+2\sqrt{x^2-4x+1}=6\sqrt{x}\)
\(\Leftrightarrow2x+2-5\sqrt{x}+2\sqrt{x^2-4x+1}-\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)
\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}}\right)=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\Leftrightarrow...\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)^2:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\sqrt{x}}=\dfrac{x-1}{x}\)
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) , dấu đẳng thức xảy ra khi và chỉ khi a = b
Ta có : \(M=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\ge\frac{4}{\sqrt{1+x^2}+\sqrt{1+y^2}}\)
Mặt khác, theo bđt Bunhiacopxki : \(\left(1.\sqrt{1+x^2}+1.\sqrt{1+y^2}\right)^2\le\left(1^2+1^2\right)\left(2+x^2+y^2\right)\)
\(\Rightarrow\sqrt{1+x^2}+\sqrt{1+y^2}\le\sqrt{20}=2\sqrt{5}\)
Do đó : \(M\ge\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}\). Dấu đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}x^2+y^2=8\\\sqrt{1+x^2}=\sqrt{1+y^2}\end{cases}\Leftrightarrow}x=y=2\)(vì x,y >0)
Vậy \(MinM=\frac{2\sqrt{5}}{5}\Leftrightarrow x=y=2\)
Điều kiện x \(\ge\frac{1}{4}\)
Đặt a = \(\sqrt{x-\frac{1}{4}}\)(a \(\ge0\))
=> x = a2 + \(\frac{1}{4}\)
=> PT <=> 2a2 + \(\frac{1}{2}\)+ \(\sqrt{a^2+\frac{1}{4}+a}\)= 2
<=> \(\sqrt{a^2+\frac{1}{4}+a}\)= \(\frac{3}{2}-2a\)
<=> a2 + 0,25 + a = 4a4 + 2,25 - 6a2
<=> 4a4 - 7a2 - a + 2 = 0
<=> (a + 1)(2a - 1)(2a2 - a - 2) = 0
<=> a = 0,5
<=> x = 0,5
\(P=\dfrac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}\cdot1+1}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}\)
=-1
\(A=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{x-x+1}=-2\sqrt{x-1}\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\) (ĐK: \(x\ge0;x\ne9\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\text{x}-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left(\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
ĐKXĐ: \(x>0\)
Áp dụng BĐT Cauchy cho 2 số dương:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}\right)^2=1\Leftrightarrow x=1\left(tm\right)\)