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a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)
a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3=3\)
hay x=0
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)ĐKXĐ : \(x>1\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)
\(A=\frac{x+2}{\sqrt{x}}\)
\(a)\)\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\right)+\left(\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\right)\)
\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\)
\(b)\) Áp dụng Cosi với hai số dương ta có :
\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}=\frac{2}{\sqrt{x}}\)
\(\Leftrightarrow\)\(x=2\)
Vậy GTNN của \(A\) là \(2\sqrt{2}\) khi \(x=2\)
Chúc bạn học tốt ~
PS : mới lớp 8 ko chắc nhé :v
a) ĐKXĐ: \(x\ne9\)
\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2x+12\sqrt{x}-18-x-5\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x\sqrt{x}-3x+12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(x+12\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x+12}{\sqrt{x}+2}\)
b) Ta có: \(P=\frac{x+12}{\sqrt{x}+2}=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)
\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\)
\(\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)
P = 4 thì \(\left(\sqrt{x}+2\right)^2=16\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
Vậy GTNN của P là 4 khi x = 4.
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)
\(=\frac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)\)
\(=x+\sqrt{x}-2\sqrt{x}-1=x-\sqrt{x}-1\)
\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)
\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)