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1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
B =\(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) + \(\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)- \(\frac{\sqrt{x}+3}{\sqrt{x}-2}\)( \(x\ge0\); \(x\ne2;3\))
= \(\frac{2\sqrt{x}-9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, B = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)= \(\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\)= \(1+\frac{4}{\sqrt{x}-3}\)
để B có gtri nguyên thì \(\frac{4}{\sqrt{x}-3}\)phải nguyên
\(\Rightarrow\left(\sqrt{x}-3\right)\varepsilonƯ\left(4\right)\)
\(\Rightarrow\left(\sqrt{x}-3\right)\varepsilon\left\{1;-1;2;-2;4;-4\right\}\)
ta có bảng sau
\(\sqrt{x}-3\) 1 -1 2 -2 4 -4
\(\sqrt{x}\) 4 2 5 1 7 -1 (L)
x 16 4 25 1 49
vậy x \(\varepsilon\){ 16 ; 4 ; 25; 1 ; 49 }
#mã mã#