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a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
a/ \(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
+ Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
+ Với \(\sqrt{x}+1=-1\Rightarrow\sqrt{x}=-2\left(vn\right)\)
+ Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)
+ Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)
Vậy x = 0 thì P nguyên
a) \(P=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)
\(=\frac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)
c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0
Vậy Min P = 1/2 <=> x = 0
\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(A=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
và \(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{x}+1}\)
\(B=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(B=\sqrt{3}+2+\frac{1}{\sqrt{3}-\sqrt{2}}+\sqrt{2}\)
\(B=\sqrt{3}+\sqrt{2}+\frac{1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{3-2+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
\(\sqrt{x}=\frac{2}{\sqrt{3}-\sqrt{2}}+3\)
\(\sqrt{x}=\frac{2\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+3\)
\(\sqrt{x}=2\sqrt{3}+2\sqrt{2}+3\)
tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D
\(C=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}}{\sqrt{x}+1}\)
P/s tham khảo nha
a) Với \(x\ge0;x\ne1\), ta có :
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(P=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)
\(P=[\frac{x-2\sqrt{x}+\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}].\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
Vậy : \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) Ta có : P > 0
\(\Leftrightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}-1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}< 1\end{cases}\Leftrightarrow}}\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)
Kết hợp với đk đề bài , ta được 0 < x < 1
Vậy với 0 < x < 1 thì P > 0
c) Với \(x=7-4\sqrt{3}=3-2.2.\sqrt{3}+4=\left(\sqrt{3}-2\right)^2\)thì :
\(P=-\sqrt{\left(\sqrt{3}-2\right)^2}\left(\sqrt{\left(\sqrt{3}-2\right)^2}-1\right)\)
\(P=-|\sqrt{3}-2|\left(|\sqrt{3}-2|-1\right)\)
\(P=\left(\sqrt{3}-2\right)\left(1-\sqrt{3}\right)\)
\(P=\sqrt{3}-3-3+2\sqrt{3}\)
\(P=3\sqrt{3}-5\)
Vậy với \(x=7-4\sqrt{3}\)thì \(P=3\sqrt{3}-5\)
d) Ta có \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Nhận thấy : \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu " = " xảy ra khi và chỉ khi
\(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì max P là \(\frac{1}{4}\)
Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)
Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)
Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)
do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)
Đến đây xét từng TH là ra
rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)
=\(1+\frac{5}{\sqrt{x}+1}\)
Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)
Đến đây thì ez rồi
Trả lời:
\(P=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\left(ĐK:x\ge0;x\ne1\right)\)
+) P > 0
\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}>0\)
\(\Leftrightarrow2\sqrt{x}-1>0\) ( vì \(\sqrt{x}+1>0\) )
\(\Leftrightarrow2\sqrt{x}>1\)
\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)
\(\Leftrightarrow x>\frac{1}{4}\)
Vậy để P > 0 thì \(x>\frac{1}{4}\) và \(x\ne1\)
+) P < 1
\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\)
\(\Leftrightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+1}-1< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Vậy để P < 1 thì \(0\le x< 4\) và \(x\ne1\)