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Câu 3:
a: =>|2x-1|=4
=>2x-1=4 hoặc 2x-1=-4
=>x=-3/2 hoặc x=5/2
b: \(\Leftrightarrow2\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}=5\)
=>3căn x+1=5
=>x+1=25/9
=>x=16/9
C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)
(đkxđ: \(c\ge0,c\ne4\))
Ta có \(A=\left(\frac{\sqrt{c}}{\sqrt{c}+2}-\frac{\sqrt{c}}{\sqrt{c}-2}+\frac{4\sqrt{c}-1}{c-4}\right).\left(\sqrt{c}+2\right)\)
\(=\frac{\sqrt{c}\left(\sqrt{c}-2\right)-\sqrt{c}\left(\sqrt{c}+2\right)+4\sqrt{c}-1}{\left(\sqrt{c}+2\right)\left(\sqrt{c}-2\right)}\left(\sqrt{c}+2\right)\)
\(=\frac{c-2\sqrt{c}-c-2\sqrt{c}+4\sqrt{c}-1}{\left(\sqrt{c}-2\right)}\)
\(=\frac{1}{2-\sqrt{c}}\)