Cho biểu thức:
P = \(\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\).
a) Rút gọn P.
b) Tìm x để P = \(\dfrac{1}{2}\).
c) Tìm GTNN của P.
Cho biểu thức P = \(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{2-\sqrt{x}}\) (với x>0; x\(\ne\)0)
a,Rút gọn biểu thức P và tìm x để P = \(\dfrac{-3}{5}\)
b,Tìm GTNN của biểu thức A=P . \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
Cho biểu thức A= \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Rút gọn biểu thức A
b, Tìm các giá trị để \(\dfrac{P}{A}\left(x-1\right)=0\)
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
cho biểu thức P = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{3-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a, rút gọn P
b, tìm x để P < \(\dfrac{1}{2}\)
c, tìm giá trị nhỏ nhất của P
Cho biểu thức :
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(1+\dfrac{2}{\sqrt{x}-1}\right)\)
a. Rút gọn C
b. Tìm x để C < -1
a/ \(B=(\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-6}{x-9}):(1+\dfrac{6}{x-9})\)
= \((\dfrac{2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-9}{x-9}+\dfrac{6}{x-9})\)
=\((\dfrac{2(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)
=\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\dfrac{x-3}{x-9})\)
=\((\dfrac{2\sqrt{x}+6+\sqrt{x}-6}{x-9}).(\dfrac{x-9}{x-3})\)
= \(\dfrac{3\sqrt{x}}{x-3}\)
Vậy B=\(\dfrac{3\sqrt{x}}{x-3}\)
b/ Để B≥0 thì \(\dfrac{3\sqrt{x}}{x-3} \)≥0
\(<=>\begin{cases} x-3 không= 0\\ 3\sqrt{x}>/0 \end{cases} \)
<=> \(\begin{cases} x không= 3\\ x>/0 \end{cases} \)
Vậy để B≥0 thì x không = 3 và x≥0
a) Ta có: \(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(1+\dfrac{2}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để C<-1 thì C+1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow2\sqrt{x}-1< 0\)
\(\Leftrightarrow x< \dfrac{1}{4}\)
Kết hợp ĐKXĐ, ta được: \(0< x< \dfrac{1}{4}\)
Cho biểu thức:
P = \(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne4,x\ne9\right)\)
a) Rút gọn P
b) Với \(x>9\), tìm GTNN của P
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
cho 2 biểu thức :
\(A=\dfrac{\sqrt{x}+2}{1-\sqrt{x}};B=\left(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
1, Rút gọn B
2, Đặt P=A.B
Tìm x ∈ Z .Tìm GTNN của P
1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{-x}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)
Cho biểu thức
P =\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
a) Tìm ĐKXĐ và rút gọn P
b) Tìm các giá trị của x để P>0
c) Tìm x để P =6
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để P>0 thì x>1
cho biểu thức Q=\(\left(\dfrac{1}{\sqrt{X}-1}-\dfrac{1}{\sqrt{X}}\right):\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-2}-\dfrac{\sqrt{X}+2}{\sqrt{X-1}}\right)\)
a rút gọn Q
b tìm x để Q>0
b) Q > 0
⇔ \(\dfrac{\sqrt{\text{x}}-2}{3\sqrt{\text{x}}}\) > 0
Do \(\text{3}\sqrt{\text{x}}>0\) ∀x⩾0
⇒ \(\sqrt{\text{x}}-2>0\)
⇔ \(\sqrt{\text{x}}>2\)
⇔ x > 4
Vậy x > 4 thì Q > 0
cho biểu thức P= \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\times\left(x-3\sqrt{x}+2\right)\)với x>0 và x≠4.
a) Rút gọn P,
b)Tìm x để P< \(\dfrac{1}{2}\)
c, Tìm gt nguyên của x để P có gt nguyên
a) đk: \(x\ne0;4\); \(x>0\)
P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để P < \(\dfrac{1}{2}\)
<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)
<=> \(\sqrt{x}< 2\)
<=> x < 4
<=> 0 < x < 4
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
cho biểu thức P =\(\left(\dfrac{\sqrt{x}}{\sqrt{x-1}}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)với 0<x≠1.
a) Rút gọn P.
b)Tìm x để P >2
`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`
`đk:x>0,x ne 1`
`P=((x+sqrtx+sqrtx)/(x-1)):(2/x+(x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):((2sqrtx+2+x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):(x+2sqrtx)/(x(sqrtx+1))`
`=(x+2sqrtx)/(x-1)*(x(sqrtx+1))/(x+2sqrtx)`
`=(x(sqrtx+1))/((sqrtx-1)(sqrtx+1))`
`=x/(sqrtx-1)`
`b)P>2`
`<=>x/(sqrtx-1)-2>0`
`<=>(x-2sqrtx+2)/(sqrtx-1)>0`
`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`
`<=>sqrtx-1>0`
`<=>x>1`
a) đk: x>0;x khác 1;0
P = \(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b) Để P > 2
<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)
<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
<=> \(\sqrt{x}-1>0\)
<=> x > 1