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Sửa đề: \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Bài này chắc là quy đồng full quá nhỉ?
a)P\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P =1/4 \(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow\sqrt{x}-2=\frac{3}{4}\sqrt{x}\)
\(\Leftrightarrow\frac{1}{4}\sqrt{x}=2\Leftrightarrow\sqrt{x}=8\Rightarrow x=64\left(TM\right)\)
P/s: Ko chắc..
Sửa đề :
a) \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Để P \(=\frac{1}{4}\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\left(TM\right)\)
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
a/ \(P=\left(\frac{\left(2+\sqrt{x}\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(P=\frac{x-4-x}{4\sqrt{x}-4}=\frac{-1}{\sqrt{x}-1}\)
b/ Thay x=\(3+2\sqrt{2}\) vào P có:
\(P=\frac{-1}{\sqrt{3+2\sqrt{2}}-1}=\frac{-1}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}=\frac{-1}{2+1-1}=\frac{-1}{2}\)
c/ \(P=\frac{1}{2}\Leftrightarrow\frac{-1}{\sqrt{x}-1}=\frac{1}{2}\Leftrightarrow-2=\sqrt{x}-1\Leftrightarrow\sqrt{x}=-1\left(vl\right)\)
Vậy ko tồn tai x để P=1/2
a) M= \(\left(1-\frac{4\sqrt{x}}{x-1}+\frac{1}{\sqrt{x}-1}\right):\frac{x-2\sqrt{x}}{x-1}\) (x > 0; x ≠ 1; x ≠ 4)
= \(\left(\frac{x-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\)\(\frac{x-1}{x-2\sqrt{x}}\)
= \(\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)
=\(\frac{x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
b)M=\(\frac{1}{2}\)
⇔ \(\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{1}{2}\)
⇔ \(2\left(\sqrt{x}-3\right)=\sqrt{x}-2\)
⇔ \(2\sqrt{x}-6=\sqrt{x}-2\)
⇔ \(2\sqrt{x}-6-\sqrt{x}-2=0\)
⇔ \(\sqrt{x}-8=0\)
⇔ \(\sqrt{x}=8\)
⇔ x = 64 (TM)
Vậy x= 64 để M=\(\frac{1}{2}\)