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a, Với \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)
Thay x = 4 => \(\sqrt{x}=2\)vào P ta được :
\(\frac{1-4}{2}=-\frac{3}{2}\)
c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)
\(\Rightarrow-x< -1\Leftrightarrow x>1\)
\(A=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}+1}{x-1}\)
\(A=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
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
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
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/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
Và: \(x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=> \(P=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}\)
=> \(P=\frac{2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)
b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\) => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)
=> \(P=-\left(2+\sqrt{3}\right)\)
c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)
<=> \(\frac{1}{\sqrt{x}-1}=-2\)
<=> \(1=-2\sqrt{x}+2\)
<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)
\(a)\)\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}-x}{1-\sqrt{x}}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(\sqrt{x}-x\sqrt{x}\right)+\left(1-x\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(1-x\right)\left(1+\sqrt{x}\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\frac{\left(x-1\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}=\frac{-\left(1-x\right)\left(1-\sqrt{x}\right)}{1-x}=\sqrt{x}-1\)
\(b)\)\(P=\sqrt{9+4\sqrt{2}}-1=\sqrt{8+4\sqrt{2}+1}-1=\sqrt{\left(2\sqrt{2}+1\right)^2}-1=2\sqrt{2}\)
\(c)\) 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 \(x=0\)