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1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
\(\text{a) }\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\sqrt{xy}\)
\(\text{b) }\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\text{c) }\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}\\ =\dfrac{\sqrt{y}-1}{x-1}\)
a)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{x}\sqrt{y}+y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}+y\)
\(=x+\sqrt{xy}+y-x+2\sqrt{xy}+y\)
\(=3\sqrt{xy}+2y\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
Bài 1:
A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-2}=-1\)
\(\Leftrightarrow\sqrt{x}-2=-1\)
\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy x = 1
Bài 2: ĐK: \(x\ge0\)
Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)
Bài 3:
a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(C=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)
Vậy MaxC = 2 khi x = 0
Còn cái GTNN chưa tính ra được, để sau nha
Bài 4: ĐK: \(x\ge0,x\ne1\)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\sqrt{x}-1\)
\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)
\(D=x-3\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)
Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)
Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)
để mk xữa đề rồi giải luôn coi có đúng o nha NGUYEN THI DIEP
xữa đề rồi giải a): \(P=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right).\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\) đk : \(\left(x\ge0;x\ne1\right)\)
\(P=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\left(1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(P=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)=1-x\)
b) ta có : \(P=\sqrt{x}\Leftrightarrow1-x=\sqrt{x}\Leftrightarrow x+\sqrt{x}-1=0\)
đặc \(\sqrt{x}=a\) \(\Rightarrow\) ta có phương trình \(\Leftrightarrow a^2+a-1=0\) \(\left(đk:x\ge0\right)\)
\(\Delta=\left(1\right)^2-4.1.\left(-1\right)=1+4=5>0\)
\(\Rightarrow\) phương trình có 2 ngiệm phân biệt
\(a_1=\dfrac{-1+\sqrt{5}}{2}\) (tmđk)
\(a_2=\dfrac{-1-\sqrt{5}}{2}\) (loại)
ta có : \(\sqrt{x}=a=\dfrac{-1+\sqrt{5}}{2}\Rightarrow x=\left(\dfrac{-1+\sqrt{5}}{2}\right)^2=\dfrac{3-\sqrt{5}}{2}\)
vậy \(x=\dfrac{3-\sqrt{5}}{2}\) thì \(P=\sqrt{x}\)
a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b: ĐểP<15/4 thì P-15/4<0
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)
=>12 căn +32-15 căn x+30<0
=>-3 căn x<-62
=>căn x>62/3
=>x>3844/9
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
a: Khi x=25 thì \(A=\dfrac{5-2}{5-1}=\dfrac{3}{4}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}=\dfrac{x-4}{x-1}\)
c: \(P=\dfrac{A}{B}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
P<1/2
=>P-1/2<0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(x\in\varnothing\)