Cho A=\(\sqrt{x}-3+\frac{10-x}{\sqrt{x}+3}\) với x≥0
a) Rút gọn A
b) Tìm GTLN của A
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a) Điều kiện xác định : \(x\ge0;x\ne1\)
\(P=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)
b) Ta có : \(P=\frac{7-3\sqrt{x}}{\sqrt{x}+4}=\frac{-3\left(\sqrt{x}+4\right)+19}{\sqrt{x}+4}=\frac{19}{\sqrt{x}+4}-3>-3\)
c) Theo b) : \(P=\frac{19}{\sqrt{x}+4}-3\)
Ta có : \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+4\ge4\Leftrightarrow\frac{19}{\sqrt{x}+4}\le\frac{19}{4}\Leftrightarrow\frac{19}{\sqrt{x}+4}-3\le\frac{7}{4}\)
\(\Rightarrow P\le\frac{7}{4}\) . Dấu "=" xảy ra khi x = 0
Vậy P đạt giá trị lớn nhất bằng \(\frac{7}{4}\) , khi x = 0
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1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
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a) \(A=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3}{\sqrt{x}+3}\)
b) \(A=\frac{1}{3}=>\frac{3}{\sqrt{x}+3}=\frac{1}{3}\)
\(=>\sqrt{x}+3=9\)
\(=>\sqrt{x}=6=>x=36\)
c) \(A\)\(lớn\)\(nhất\)\(< =>\frac{3}{\sqrt{x}+3}lớn\)\(nhất\)
\(=>\sqrt{x}+3\)\(nhỏ\)\(nhất\)
\(Mà\)\(\sqrt{x}+3>=3
\)
\(Do\)\(đó\)\(\sqrt{x}+3=3=>x=0\)
\(a.A=\sqrt{x}-3+\frac{10-x}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{10-x}{\sqrt{x}+3}=\frac{x-9+10-x}{\sqrt{x}+3}=\frac{1}{\sqrt{x}+3}=\frac{\sqrt{x}-3}{x-9}\)
\(b.\)Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{1}{\sqrt{x}+3}\ge\frac{1}{3}\forall x\)
Vậy \(A_{Min}=\frac{1}{3}\Leftrightarrow x=0\)