RUT GON:
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
Những câu hỏi liên quan
RUT GON BIEU THUC \(\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}\)
nhan ca tu va mau voi\(\sqrt{2}\) ta dc
\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)
=\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
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rut gon A= \(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\) - \(\frac{2x+\sqrt{x}}{\sqrt{x}}\)+\(\frac{2x-2}{\sqrt{x}-1}\)
rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
RUT GON:
\(\sqrt{2x+1-2\sqrt{\left(x-1\right)\left(x+2\right)}}\)
ĐKXĐ: \(x\ge1\)
\(=\sqrt{x+2-2\sqrt{\left(x-1\right)\left(x+2\right)}+x-1}\)
\(=\sqrt{\left(\sqrt{x+2}-\sqrt{x-1}\right)^2}\)
\(=\sqrt{x+2}-\sqrt{x-1}\)
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rut gon roi tinh gia tri cua x de A=-1
\(\sqrt{2}-\sqrt{x+2\sqrt{2x-4}}\)
\(A=\sqrt{2}-\sqrt{x+2\sqrt{2x-4}}\) ( ĐKXĐ: \(x\ge2\))
\(\Rightarrow A\sqrt{2}=2-\sqrt{2x+4\sqrt{2x-4}}\)
\(=2-\sqrt{\left(\sqrt{2x-4}+2\right)^2}\)
\(=2-\sqrt{2x-4}-2\)
\(=-\sqrt{2x-4}\)
\(\Rightarrow A=-\sqrt{\frac{2x-4}{2}}\)
\(=-\sqrt{x-2}\)
\(A=-1\Leftrightarrow-\sqrt{x-2}=-1\)
\(\Leftrightarrow\sqrt{x-2}=1\)
\(\Leftrightarrow x=3\)( Thỏa mãn ĐKXĐ )
TK NHA!
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Cho A=\(4x-\sqrt{8}-\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
a) Rut gon A khi x >2
b)Tinh A khi x=\(\sqrt{-2}\)
câu b) \(\sqrt{-2}\) không xác định
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a) \(A=4x-\sqrt{8}-\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}-\frac{\sqrt{x^2\left(x+2\right)}}{\sqrt{x+2}}=4x-\sqrt{8}-x=3x-\sqrt{8}\)
b) \(x=\sqrt{-2}\) (không thỏa mãn)
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rut gon \(y=\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x-1}}{x-\sqrt{x}}-\frac{^{x^2}+\sqrt{x}}{x\sqrt{x}+x}\)
\(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a)Rut gon P?
b)Tinh gia tri cua P voi \(x=3-2\sqrt{2}\)?
Cho A = \(\frac{3\sqrt{x}-3}{x\sqrt{x}-2x+2\sqrt{x}-1}-\frac{4x\sqrt{x}-4}{x^3-1}\)(x>1). Rut gon A va tim x de A=1