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a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
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:
\(B=4\cdot\left(-\sqrt{2}\right)+\left(-\sqrt{2}\right)^2=2-4\sqrt{2}\)
a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)
\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)
\(2\sqrt{x}-1>0\);\(4x>0\)
\(\Rightarrow x>0\)thì \(A>A^2\)
\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{x\sqrt{x}-8}{\sqrt{x}-2}-\frac{2x-\sqrt{x}}{\sqrt{x}}-5\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}-2}-\frac{\left(2x-\sqrt{x}\right)\sqrt{x}}{x}-5\)
\(\Leftrightarrow P=x+2\sqrt{x}+4-\frac{x\left(2\sqrt{x}-1\right)}{x}-5\)
\(\Leftrightarrow P=x+2\sqrt{x}+4-2\sqrt{x}+1-5\)
\(\Leftrightarrow P=x\)
\(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}+\frac{\sqrt{x^2}.\sqrt{x+2}}{\sqrt{x+2}}=4x-\sqrt{8}+|x|\)
có còn rút gọn đc nữa không nhỉ
Thanks Nghĩa nha, làm như thế là được rồi^^