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Y=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}\)-1-\(\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
=\(\sqrt{x}\left(\sqrt{x}+1\right)\)-1-(\(2\sqrt{x}+1\))
=2\(\sqrt{x}+\sqrt{x}\)-1-2\(\sqrt{x}\)-1
=\(\sqrt{x}-2\)
x>0
y=[√x.(√x+1).(x-√x+1)]/(x-√x+1)-1-[√x.(2√x+1)]
=√x.(√x+1)-2√x-2
=x-√x-2
b.
y=(√x-1/2)^2-9/4≥-9/4
x=1/4
c.
x≥4=>(√x-1/2)^2≥9/4=>y≥0
=>y≥0=>|y|=y
=>y-|y|=y-y=0
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\) = \(\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\) = \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)
= \(\dfrac{\sqrt{y-1}}{x-1}\)
Chúc bạn học tốt :3
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
a) \(ĐKXĐ:x>0\)
\(Y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-1-2\sqrt{x}-1\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x+\sqrt{x}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x-\sqrt{x}-2\)
b) Ta có \(Y=x-\sqrt{x}-2=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy \(Min_Y=-\frac{9}{4}\Leftrightarrow x=\frac{1}{4}\)
c) Để \(Y-\left|Y\right|=0\)
\(\Leftrightarrow Y=\left|Y\right|\)
\(\Leftrightarrow Y\ge0\)
\(\Leftrightarrow x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (Vì \(\sqrt{x}+1\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\) (ĐPCM)
2
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
t
sao vậy bạn