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a) ĐKXĐ: \(x\ge0\); \(1-4x\ne\)0; \(2\sqrt{x}-1\ne0\); \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\ne\)0
<=> \(x\ge0\); x \(\ne\)1/4
Ta có: \(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(A=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x+2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}\right)\)
\(A=\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{6x+4x+2\sqrt{x}}\)
\(A=\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\)
b)Với x \(\ge\)0 và x \(\ne\)1/4
Ta có: A > A2 <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\left(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)^2\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\left(1-\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)>0\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10x+2\sqrt{x}-\sqrt{x}+1}{10x+2\sqrt{x}}>0\)
<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10+\sqrt{x}+1}{10x+2\sqrt{x}}>0\)
<=> \(\sqrt{x}-1>0\) <=> \(x>1\)
c) Với x\(\ge\)0 và x \(\ne\)1/4 (1)
Ta có: \(\left|A\right|>\frac{1}{4}\) <=> \(\orbr{\begin{cases}A>\frac{1}{4}\\A< -\frac{1}{4}\end{cases}}\)
TH1: \(A>\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\frac{1}{4}\)
<=> \(4\left(\sqrt{x}-1\right)>10x+2\sqrt{x}\)
<=> \(4\sqrt{x}-4>10x+2\sqrt{x}\)
<=> \(10x-2\sqrt{x}+4< 0\)(vô liia vì \(10x-2\sqrt{x}+4>0\))
TH2: \(A< -\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}< -\frac{1}{4}\)
<=> \(4\left(\sqrt{x}-1\right)< -10x-2\sqrt{x}\)
<=> \(4\sqrt{x}-4+10x+2\sqrt{x}< 0\)
<=> \(10x+6\sqrt{x}-4< 0\)
<=> \(5x+3\sqrt{x}-2< 0\)
<=> \(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)< 0\)
<=> \(x< \frac{4}{25}\) (2)
Từ (1) và (2) => \(0\le x< \frac{4}{25}\)
1. Áp dụng Min - cốp - ski, ta được: \(\sqrt{\frac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\frac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\frac{9}{\left(c+a\right)^2}+b^2}\)\(\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{c+a}\right)^2+\left(a+b+c\right)^2}\)\(\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)(Bunyakovsky dạng phân thức)
Đặt \(t=a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)thì ta cần chứng minh: \(\sqrt{\frac{729}{4t^2}+t^2}\ge\frac{3\sqrt{13}}{2}\Leftrightarrow\frac{729}{4t^2}+t^2\ge\frac{117}{4}\)\(\Leftrightarrow\frac{\left(t+3\right)\left(t-3\right)\left(2t+9\right)\left(2t-9\right)}{4t^2}\ge0\)*đúng bởi \(t-3\le0;t+3>0;2t+9>0;2t-9< 0;4t^2>0\)*
Đẳng thức xảy ra khi t = 3 hay a = b = c = 1
2. Ta có: \(\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}-3=\frac{\left(x^2-y^2\right)^2\left(x^4+y^4+x^2y^2\right)}{x^2y^2\left(x^2+y^2\right)^2}\ge0\)\(\Rightarrow\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge3\)
Đẳng thức xảy ra khi x = y
a) A = \(\sqrt{-x^2+x+\dfrac{3}{4}}=\sqrt{1-\left(x-\dfrac{1}{2}\right)^2}\le\sqrt{1}=1\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))
Vậy max A = 1 (khi và chỉ khi x = \(\dfrac{1}{2}\))
b) B = \(\sqrt{\left(2x^2-x-1\right)^2+9}\ge\sqrt{9}=3\) (dấu "=" xảy ra \(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=1;x=-\dfrac{1}{2}\)).
Vậy min B = 3 (khi và chỉ khi x = 1 hoặc x = \(-\dfrac{1}{2}\))
c) C = \(\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\);
C \(\ge\left|2-5x+5x\right|=\left|2\right|=2\) (dấu "=" xảy ra \(\Leftrightarrow\left(2-5x\right).5x\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2-5x\ge0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x\le0\\2-5x\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le x\le\dfrac{2}{5}\)).
Vậy min C = 2 (khi và chỉ khi \(0\le x\le\dfrac{2}{5}\))
\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)
\(2\sqrt{x}-1>0\);\(4x>0\)
\(\Rightarrow x>0\)thì \(A>A^2\)
b)\(\sqrt{25x^2}=19\)
\(\Leftrightarrow5x=19\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
c)\(\sqrt{x-7}+3=0\)
\(\Leftrightarrow\sqrt{x-7}=-3\)
\(\Leftrightarrow x-7=9\)
\(\Leftrightarrow x=16\)