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ĐKXĐ: ...
Đặt \(\sqrt{2x-1}=t\ge0\Rightarrow x=\frac{t^2+1}{2}\)
\(\Rightarrow A=\frac{2t^2+6t+4}{t^2+4t+3}=\frac{2\left(t+1\right)\left(t+2\right)}{\left(t+1\right)\left(t+3\right)}=\frac{2\left(t+2\right)}{t+3}=2-\frac{2}{t+3}\ge2-\frac{2}{3}=\frac{4}{3}\)
Dấu "=" xảy ra khi \(t=0\Leftrightarrow x=\frac{1}{2}\)
đkxđ \(x\ne1;x\ge0\)
\(P=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2}{\left(\sqrt{x}\right)^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{1}{\sqrt{x}-1}-\frac{x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(P=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}+1-x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}+2}{\left(\sqrt{x}\right)^3-1}\)
Giải:
ĐKXĐ của P là \(x\ge2\)và \(x\ne5\)
Phân tích tử:
x-5 = x-2-3
= (\(\sqrt{x-2}\)-\(\sqrt{3}\))(\(\sqrt{x-2}\)+\(\sqrt{3}\))
Xét P=\(\frac{\left(\sqrt{x-2}-\sqrt{3}\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{\sqrt{x-2}-\sqrt{3}}\)
= \(\sqrt{x-2}+\sqrt{3}\)
=> Min P= \(\sqrt{3}\)khi X=2.
Mình chỉ có thể tìm GTNN, còn GTLN thì mk chịu.
\(A=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}=x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\left(x\ge0\right)\)
\(\Rightarrow A_{Min}=-\dfrac{1}{4}."="\Leftrightarrow x=\dfrac{1}{4}\left(TM\right)\)
đkxđ:x>=0
\(A^2=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x+1}\right)^2}=\frac{x-2\sqrt{x}+1}{x+1}=1-\frac{2\sqrt{x}}{x+1}\)
vì \(\left(\sqrt{x}-1\right)^2=x-2\sqrt{x}+1>=0\Rightarrow x+1>=2\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x+1}< =\frac{x+1}{x+1}=1\Rightarrow1-\frac{2\sqrt{x}}{x+1}>=1-1=0\)
dấu = xảy ra khi x=1
vậy min A là 0 khi x-=1
a: \(P=\dfrac{\left[\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-4+2\left(\sqrt{x}+1\right)\right]}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+\sqrt{x}-2\sqrt{x}-4+2\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c: Để |P|>P thì P<0
\(\Leftrightarrow\sqrt{x}-1< 0\)
hay 0<x<1
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
\(\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\sqrt{x}+1-4}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+1}-\frac{4}{\sqrt{x}+1}=1-\frac{4}{\sqrt{x+1}}\)
Để \(1-\frac{4}{\sqrt{x}+1}\) lớn nhất <=> \(\frac{4}{\sqrt{x}+1}\) lớn nhất => \(\sqrt{x}+1\)nhỏ nhất
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\)
Dấu "=" xảy ra <=> \(\sqrt{x}=0\Rightarrow x=0\)
Vậy .........
ĐK: x > 0
a) Rút gọn M
M = \(\frac{\sqrt{x}}{x+\sqrt{x}}:\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)
=> M \(\le\)1/3
=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn
Vậy max M = 1/3 tại x = 1
