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Ta có: P = \frac{4\sqrt{x}}{8x} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} - 2} : \frac{\sqrt{x} + 2}{x - 4} \cdot \frac{\sqrt{x} - 2}{\sqrt{x} + 2} = \frac{4\sqrt{x}(\sqrt{x} + 2)}{(8x)(\sqrt{x} - 2)} : \frac{x - 4}{x - 4} = \frac{4(\sqrt{x} + 2)}{8(\sqrt{x} - 2)} = \frac{1}{\sqrt{x} - 2} 2) Tìm các giá trị của x để P = -4: Ta có: P = -4 \Rightarrow \frac{1}{\sqrt{x} - 2} = -4 \Rightarrow \sqrt{x} - 2 = -\frac{1}{4} \Rightarrow \sqrt{x} = \frac{7}{4} \Rightarrow x = \left(\frac{7}{4}\right)^2 = \frac{49}{16} Vậy x = 49/16 là giá trị cần tìm.
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=\(\frac{1}{2}\)
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
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 \(=\frac{1}{4}\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\left(TM\right)\)