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b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}\)
\(=4x-\sqrt{8}+\frac{x\left(x+2\right)}{x+2}\)
\(=4x-\sqrt{8}+x\)
\(=5x-\sqrt{8}\)
Với \(x=-\sqrt{2}\) ta có:
\(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)
a)\(\sqrt{\frac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\frac{x^2-1}{x-3}=\frac{\sqrt{\left(x-2\right)^4}}{\sqrt{\left(3-x\right)^2}}+\frac{x^2-1}{x-3}=\frac{\left(x-2\right)^2}{x-3}+\frac{x^2-1}{x-3}=\frac{x^2-4x+4+x^2-1}{x-3}=\frac{2x^2-4x+3}{x-3}\)
Tại x=0,5 thay vào ta có:
\(A=\frac{2\cdot\left(0,5\right)^2-4\cdot0,5+3}{0,5-3}=-\frac{3}{5}\)
b)\(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2}\cdot\sqrt{x+2}}{\sqrt{x+2}}\)\(=4x-\sqrt{8}+x^2\)
Tại \(x=-\sqrt{2}\) thay vào ta có:
\(B=4\cdot\left(-\sqrt{2}\right)+\left(-\sqrt{2}\right)^2=2-4\sqrt{2}\)
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(=1-\left(\frac{2\left(1-2\sqrt{x}\right)+5\sqrt{x}-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}\right):\frac{\sqrt{x}-1}{\left(1+2\sqrt{x}\right)^2}\)
\(=1-\frac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}.\frac{\left(1+2\sqrt{x}\right)^2}{\sqrt{x}-1}=1-\frac{1+2\sqrt{x}}{1-2\sqrt{x}}=2-\frac{2}{1-2\sqrt{x}}\)
để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Lời giải:
\(M=\frac{\sqrt{x-2\sqrt{2}}}{\sqrt{x^2-4x\sqrt{2}+8}}-\frac{\sqrt{x+2\sqrt{2}}}{\sqrt{x^2+4x\sqrt{2}+8}}\)
\(=\frac{\sqrt{x-2\sqrt{2}}}{\sqrt{(x-2\sqrt{2})^2}}-\frac{\sqrt{x+2\sqrt{2}}}{\sqrt{(x+2\sqrt{2})^2}}=\frac{\sqrt{x-2\sqrt{2}}}{x-2\sqrt{2}}-\frac{\sqrt{x+2\sqrt{2}}}{x+2\sqrt{2}}\)
\(=\frac{1}{\sqrt{x-2\sqrt{2}}}-\frac{1}{\sqrt{x+2\sqrt{2}}}\)
Thay $x=3$:
\(M=\frac{1}{\sqrt{3-2\sqrt{2}}}-\frac{1}{\sqrt{3+2\sqrt{2}}}=\frac{1}{\sqrt{2-2\sqrt{2}+1}}-\frac{1}{\sqrt{2+2\sqrt{2}+1}}\)
\(=\frac{1}{\sqrt{(\sqrt{2}-1)^2}}-\frac{1}{\sqrt{(\sqrt{2}+1)^2}}=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=\frac{2}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{2}{2-1}=2\)