(Đề thi tuyển sinh vào 10 - chuyên Thái Bình)
Cho biểu thức
\(A=\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\) với \(x>0,\text{ }x\ne1\).
a) Rút gọn $A$.
b) Cho \(B=\left(x-\sqrt{x}+1\right).A\) . Chứng minh rằng \(B>1.\)
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 )
b) \(B=\left(x-\sqrt{x}+1\right).A\)
Với \(x>0.x\ne1\)thì \(B=\left(x-\sqrt{x}+1\right).\frac{1}{\sqrt{x}}=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\)
\(B=\left(\sqrt{x}+\frac{1}{\sqrt{x}}-2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}\right)+2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}-1=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+2-1\)
\(B=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1\)
Ta có:
\(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2\ge0\forall x>0\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}=0\Leftrightarrow\sqrt[4]{x}=\frac{1}{\sqrt[4]{x}}\Leftrightarrow x=1\)(thỏa mãn điều kiện x>0)
Mà theo đề bài, \(x\ne1\)nên dấu bằng không xảy ra
Do đó : \(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2>0\forall x\left(x>0;x\ne1\right)\)
\(\Rightarrow\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1>1\forall x\left(x>0;x\ne1\right)\)
\(\Rightarrow B>1\forall x\left(x>0;x\ne1\right)\)
Vậy với \(x>0;x\ne1\)thì \(B>1\)
a, \(A= \dfrac{4\sqrt{x}+2}{(\sqrt{x}+1)4\sqrt{x}}\)
A=\(\dfrac{1}{\sqrt{x}}\)
a, A=\(\dfrac{1}{\sqrt{x}}\) với x>\(0;x\ne1\) b,B=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\) >1 \(\Rightarrow\dfrac{x-\sqrt{x}+1-\sqrt{x}}{\sqrt{x}}>0\) \(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\Rightarrow\left(\sqrt{x}-1\right)^2\) > 0\(\Rightarrow x>1\) hay B>1 Vậy B>1 \(\forall x>0;x\ne1\)
a/ ĐKXD : x > 0 , x≠1
A= \(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(x-1\right)}\right).\left(\dfrac{x+2\sqrt{x}+1}{4\sqrt{x}}-1\right)\)=\(\dfrac{\sqrt{x}-1+3\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(x-1\right)}.\dfrac{x-2\sqrt{x}+1}{4\sqrt{x}}=\dfrac{4.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)=\(\dfrac{1}{\sqrt{x}}\)
Vậy A=\(\dfrac{1}{\sqrt{x}}\) với x > 0 , x≠1
b/ Cho B=\(\left(x-\sqrt{x}+1\right).\dfrac{1}{\sqrt{x}}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}\)
≥\(2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}\)=2
⇒\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\)-1≥1
hay B≥1 . Dấu "=" xảy ra ⇔ \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\)⇒x=1 (TMDK)
Vậy Bmin=1 khi x=1
a, A = \(\dfrac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(x-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
A = \(\dfrac{1}{\sqrt{x}}\)
b, Ta có :
\(\Rightarrow B=\left(x-\sqrt{x}-1\right).A\)
\(\Rightarrow B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
Ta có:
\(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}>1\)
\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>1\)
Mà \(\left(\sqrt{x}-1\right)^2>1\)
\(\Rightarrow x>1\left(B>1\right)\)
Vậy \(B>1\) với mọi x >0, x \(\ne1\)
Với x >0 ; x ≠1, ta có
A=\([\dfrac{1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{3\sqrt{x}+5}{(\sqrt{x}-1)^2(\sqrt{x}+1)}]\times\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{4\sqrt{x}}\)
A=\(\dfrac{\sqrt{x}-1+3\sqrt{x}+5}{(\sqrt{x}-1)^2(\sqrt{x}+1)}\times\dfrac{(\sqrt{x}-1)^2}{4\sqrt{x}}\)
A=\(\dfrac{4(\sqrt{x}+1)(\sqrt{x}-1)^2}{4(\sqrt{x}-1)^2(\sqrt{x}+1)\sqrt{x}}\)
A=\(\dfrac{1}{\sqrt{x}}\)
Vậy với x >0 ;x≠1 thì A=\(\dfrac{1}{\sqrt{x}}\)
b) B =\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)=\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}}\)
Áp dụng bất đẳng thức Cô si , ta có
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}}{\sqrt{x}}}\)
⇒\(\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\ge2-1=1\)
Dấu = xảy ra ⇔\(\dfrac{x+1}{\sqrt{x}}=2\Leftrightarrow2x+2=\sqrt{x}\)(Vô lý vì x>\(\sqrt{x}\) với∀ x>0,x≠1)
Vậy B>1 với mọi x>0 , x≠1
a) A= \(\dfrac{1}{\sqrt{x}}\)
b) \(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)= \(\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\)\(\ge2\sqrt{1}-1\)\(\ge1\)( Cosi)
a,\(\left[\dfrac{1}{x-1}+\dfrac{3\sqrt{5}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right].\dfrac{\left(\sqrt{x}+1\right)^x-4\sqrt{x}}{4\sqrt{x}}\)
\(=\dfrac{4}{\left(\sqrt{x}-1\right)^2}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
=\(\dfrac{1}{\sqrt{x}}\)
b,B=\(\left(x-\sqrt{x}+1\right)\dfrac{1}{\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\)
B>1
Áp dụng cô si
\(=>\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\ge1\)
Mà x\(\ne\)1
=>B>1
B>1\(\forall x>0,x\ne1\)
a) A=\(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right)\)
=\(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right).\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{4\sqrt{x}}\)
=\(\dfrac{4\sqrt{x}+4}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
=\(\dfrac{1}{\sqrt{x}}\)
b) Ta có
\(B=\left(x-\sqrt{x}+1\right).A=\left(x-\sqrt{x}+1\right).\dfrac{1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\)
Áp dụng bất đẳng thức Cô si, ta có , đẳng thức xảy ra khi .
Tuy nhiên, theo điều kiện nên .
Vậy .
a) A=\(\dfrac{1}{\sqrt{x}}\)
b) B > 1
a, A = \(\dfrac{1}{\sqrt{x}}\)với x > 0; x khác 1
b, Với x >0 và x khác 1 có :
B = \(\left(x-\sqrt{x}+1\right)A=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{^{\left(\sqrt{x}-1\right)^2+\sqrt{x}}}{\sqrt{x}}>\dfrac{\sqrt{x}}{\sqrt{x}}=1\)(ĐPCM)
( do x khác 1 => ( \(\sqrt{x}\)-1)\(^2\)>0
