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Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
a) \(M=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}-1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x^3}-1^3}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}-\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2+\sqrt{x}}{\sqrt{x}}-\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2x+2+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}+2x+2\sqrt{x}+2+x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}+2+3\sqrt{x}+3x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{2\left(x\sqrt{x}+1\right)+3\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)+3\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(3\sqrt{x}+2x+2\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{5\sqrt{x}+2x+2}{\sqrt{x}}\)
xin lỗi bn nhé, mk chỉ bt làm đến đây thôi, k bt có đúng k
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) P = \(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)\).
P = \(\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)}\)
P = \(\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1-x+\sqrt{x}}\)
P = \(\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
P = \(\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
P = \(x-1\).
b) P = \(\frac{9}{2}\).
⇔ \(x-1=\frac{9}{2}\)
⇔ \(x=\frac{11}{2}\).
Vậy \(x=\frac{11}{2}\)thì P = \(\frac{9}{2}\).
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
a: \(M=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: Để M=9/2 thì \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\dfrac{9}{2}\)
=>2x-5 căn x+2=0
=>(căn x-2)(2 căn x-1)=0
=>x=1/4 hoặc x=4
c: \(M-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>M>4
a: \(A=\left(\dfrac{\sqrt{x}}{x+2}+\dfrac{6\sqrt{x}}{x-4}\right)\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{x-2\sqrt{x}+6\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}+2}{1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\)
b: \(M=A:B=\dfrac{x+4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}}{\sqrt{x}+1}\)
b: \(M-1=\dfrac{x+4\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{x+3\sqrt{x}-1}{\sqrt{x}+1}>0\)
=>M>1
a) Ta có: \(M=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Ta có: M-1
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{-2}{\sqrt{x}}< 0\forall x\) thỏa mãn ĐKXĐ
hay M<1