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B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(A=\sqrt{a\left(a-1\right)}\)
Q=\(\frac{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}{\sqrt{x+\sqrt{2x-1}-\sqrt{x-\sqrt{2x-1}}}}\)(x\(\ge2\))
\(=\frac{\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2}{\left(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\right)\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)}\)
=\(\frac{2x+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}}{2\sqrt{2x-1}}\)
=\(\frac{x+\sqrt{x^2-2x+1}}{\sqrt{2x-1}}=\frac{x+\sqrt{\left(x-1\right)^2}}{\sqrt{2x-1}}\) \(=\frac{x+x-1}{\sqrt{2x-1}}=\frac{2x-1}{\sqrt{2x-1}}=\sqrt{2x-1}\)
vậy \(Q=\sqrt{2x-1}với\)\(x\ge2\)
~ happy new year~
Happy new year !
Cố gắng kì sau làm CTV nha chị :D