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\(E=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(=2x-1+2x-3\)
\(=4x-4\)
Làm nốt
\(\sqrt{x}+2\sqrt{1-x}\le\sqrt{\left(1+4\right)}=\sqrt{5}\)
Mà ta có điều kiện là \(0\le x\le1\)
=> E \(\ge1\)
Vậy GTLN là \(\sqrt{5}\)đạt được khi x = \(\frac{1}{5}\)
Đạt GTNN là 1 khi x = 1
ĐK: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}-3\right)+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x-6\sqrt{x}+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-9x+9}{x-9}\)
a) Thay x=25 vào biểu thức \(A=\frac{7}{\sqrt{x}+8}\), ta được:
\(A=\frac{7}{\sqrt{25}+8}=\frac{7}{5+8}=\frac{7}{13}\)
Vậy: khi x=25 thì \(A=\frac{7}{13}\)
b) Ta có: \(B=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)
c) Ta có: \(P=A\cdot B\)
\(=\frac{7}{\sqrt{x}+8}\cdot\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)
ĐKXĐ: \(x\ge0\)
Để P có giá trị nguyên thì \(7⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-7;-1;7\right\}\)
\(\Leftrightarrow\sqrt{x}+3=7\)(vì \(\sqrt{x}+3\ge3\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}=4\)
hay x=16(nhận)
Vậy: Khi x=16 thì P nguyên
d) Ta có: \(\sqrt{x}+3\ge3\forall x\ge0\)
\(\Leftrightarrow\frac{7}{\sqrt{x}+3}\le\frac{7}{3}\forall x\ge0\)
Dấu '=' xảy ra khi x=0
Vậy: Giá trị lớn nhất của biểu thức \(P=A\cdot B\) là \(\frac{7}{3}\) khi x=0
e) Để \(P=\frac{1}{2}\) thì \(\frac{7}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}+3=7\cdot2=14\)
\(\Leftrightarrow\sqrt{x}=14-3=11\)
hay x=121(nhận)
Vậy: để \(P=\frac{1}{2}\) thì x=121
Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) ( ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))
\(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)
\(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)
a) Với \(x\ge0\)và \(x\ne1\)ta có:
\(P=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)
\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{10\sqrt{x}-2x+5\sqrt{x}-3-x-5\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-\left(3x-10\sqrt{x}+7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}\)
b) \(P=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}=\frac{-3\sqrt{x}-12+19}{\sqrt{x}+4}=\frac{-3\left(\sqrt{x}+4\right)+19}{\sqrt{x}+4}=-3+\frac{19}{\sqrt{x}+4}\)
Vì \(x\ge0\); \(x\ne1\)\(\Rightarrow\sqrt{x}+4\ge4\)
\(\Rightarrow\frac{19}{\sqrt{x}+4}\le\frac{19}{4}\)\(\Rightarrow P\le-3+\frac{19}{4}=\frac{7}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow x=0\)( thỏa mãn )
Vậy \(maxP=\frac{7}{4}\)\(\Leftrightarrow x=0\)
\(\dfrac{\left(\sqrt{3-x}+\sqrt{x+1}\right)^2}{2}\le3-x+x+1=4\)\(\sqrt{3-x}+\sqrt{x+1}\le2\sqrt{2}\)
dang thuc khi \(x=1\)