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\(A=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}+1}{x-1}\)
\(A=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
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
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: 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:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
P = \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\). \(\frac{\left(x-1\right)^2}{2}\)( x\(\ge0\); x\(\ne\)1)
= \(\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\) . \(\frac{\left(x-1\right)^2}{2}\)
= \(\frac{x-\sqrt{x}+2-x-\sqrt{x}+2}{\sqrt{x}-1}\). \(\frac{x-1}{2}\)
= \(\frac{\left(-2\sqrt{x}+4\right)\left(\sqrt{x}+1\right)}{2}\)
= \(\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)\)
= -x2 + \(\sqrt{x}\)+ 2
b. tự tính nha
c, P = -x2 + \(\sqrt{x}+2\)
= - (x2 - 2.x.1/2 + 1/4) +2 +1/4
= - (x-1/2)2+ 9/4
ta có (x - 1/2)2 \(\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\forall x\)
dấu "=" xảy ra khi và chỉ khi x-1/2 = 0
x=1/2
vậy GTLN của P= 9/4 khi và chỉ khi x=1/2
#mã mã#
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
\(a)\)\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}-x}{1-\sqrt{x}}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(\sqrt{x}-x\sqrt{x}\right)+\left(1-x\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(1-x\right)\left(1+\sqrt{x}\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\frac{\left(x-1\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}=\frac{-\left(1-x\right)\left(1-\sqrt{x}\right)}{1-x}=\sqrt{x}-1\)
\(b)\)\(P=\sqrt{9+4\sqrt{2}}-1=\sqrt{8+4\sqrt{2}+1}-1=\sqrt{\left(2\sqrt{2}+1\right)^2}-1=2\sqrt{2}\)
\(c)\) Ta có : \(\frac{2}{P}=\frac{2}{\sqrt{x}-1}\)
Để P nguyên thì \(\frac{2}{\sqrt{x}-1}\) nguyên hay \(2⋮\left(\sqrt{x}-1\right)\)\(\Rightarrow\)\(\left(\sqrt{x}-1\right)\inƯ\left(2\right)\)
Mà \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)\(\Rightarrow\)\(x\in\left\{\sqrt{2};0;\sqrt{3}\right\}\)
Do x là số chính phương nên \(x=0\)
Vậy để \(\frac{2}{P}\) là số nguyên thì \(x=0\)