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A=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
ĐKXĐ :x\(\ge\)0;x\(\ne\)1;x\(\ne\)3
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x+16}{\sqrt{x}+3}\)
b, x =(\(\sqrt{2}-1)^2\)
Thay x =(\(\sqrt{2}-1)^2\)thỏa mãn đk vào a có:
A=\(\dfrac{\left(\sqrt{2}-1\right)^2+16}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
=\(\dfrac{2-2\sqrt{2}+1+16}{\sqrt{2}-1}\)
=\(\dfrac{19\sqrt{2}+19-4-2\sqrt{2}}{2-1}\)
=\(17\sqrt{2}+15\)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a: \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
b: Khi \(x=7-4\sqrt{3}\) vào P, ta được:
\(P=\dfrac{7-4\sqrt{3}+16}{2-\sqrt{3}+3}=\dfrac{23-4\sqrt{3}}{5-\sqrt{3}}\)
a: \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
b: \(P=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)
\(\Leftrightarrow P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot5-6=10-6=4\)
Dấu = xảy ra khi x=4
ĐK: x\(\ge0,x\ne1\)
a) \(Q=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Ta có \(Q=0,5\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=0,5\Leftrightarrow2-5\sqrt{x}=0,5\sqrt{x}+1,5\Leftrightarrow0,5=5,5\sqrt{x}\Leftrightarrow\sqrt{x}=\dfrac{1}{11}\Leftrightarrow x=\dfrac{1}{121}\left(tm\right)\)
Vậy \(x=\dfrac{1}{121}\) thì \(Q=0,5\)
c) Ta có \(Q=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}=\dfrac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)
Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}+\left(-5\right)\le\dfrac{2}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{2}{3}\Leftrightarrow Q\le\dfrac{2}{3}\)
Dấu bằng xảy ra khi x=0
Vậy GTLN của Q=\(\dfrac{2}{3}\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
b) A= \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\) + \(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)- \(\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
A= \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)}\) - \(\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)- \(\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}\)
= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c) GTLN (Max)
A= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
= -5+\(\dfrac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\)\(\ge\)0 (ĐKXĐ) \(\Rightarrow\) \(\sqrt{x}+3\ge3\)
\(\Rightarrow\) \(\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\)
\(\Rightarrow\) \(-5+\dfrac{17}{\sqrt{x}+3}\le-5+\dfrac{17}{3}\)
\(\Leftrightarrow\) A\(\le\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\sqrt{x}=0\) \(\Rightarrow\) \(x=0\)
Vậy Max A =\(\dfrac{2}{3}\) khi \(x=0\)
Lời giải:
ĐKXĐ: \(x\geq 0, x\neq 1\)
Ta có:
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}+3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{x+2\sqrt{x}+3}-\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{3x+7\sqrt{x}-6}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{2x+\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}+3)}\)
\(=\frac{-5x+7\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(\sqrt{x}-1)(2-5\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b)
\(A=\frac{1}{2}\Leftrightarrow \frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow 2(2-5\sqrt{x})=\sqrt{x}+3\)
\(\Leftrightarrow 1=11\sqrt{x}\Rightarrow x=\frac{1}{121}\)
c)
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{17-5(\sqrt{x}+3)}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}-5\)
Ta thấy: \(\sqrt{x}\geq 0\Rightarrow \sqrt{x}+3\geq 3\Rightarrow A=\frac{17}{\sqrt{x}+3}-5\leq \frac{17}{3}-5=\frac{2}{3}\)
Vậy \(A_{\max}=\frac{2}{3}\)
Dấu bằng xảy ra khi $x=0$
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}.\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x\left(\sqrt{x}-1\right)+16.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(\sqrt{x}-1\right).\left(x+16\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
ĐKXĐ : \(x\ge0\) ; \(x\ne1\) ; \(x\ne9\)