P=\(\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\\ \)
a, rut gon
b, tim x de P<\(\frac{15}{4}\)
c, tim gia tri lon nhat cua P
K
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BT
12 tháng 5 2017
a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)
=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)
b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)
<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:
<=> 16x=(x-1)2
<=> 16x=x2-2x+1
<=> x2-18x+1=0
\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)
=> \(x_1=9-4\sqrt{5}\)
\(x_2=9+4\sqrt{5}\)
26 tháng 7 2019
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{2}\)
\(=\frac{-\sqrt{x}}{\sqrt{x}-1}\)
Để p = -2 \(\Rightarrow\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)
\(\frac{-\sqrt{x}}{\sqrt{x}-1}=-2\)
\(\Rightarrow-\sqrt{x}=-2\left(\sqrt{x}-1\right)\)
\(\Rightarrow-\sqrt{x}=-2\sqrt{x}+2\)
\(\Rightarrow-\sqrt{x}+2\sqrt{x}=2\)
\(\Rightarrow\sqrt{x}=2\)
\(\Rightarrow x=4\)
NV
21 tháng 5 2017
ĐKXĐ: \(x\ge0\)
a/ Đề \(=\left(\frac{1-\sqrt{x}^3}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left[\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right]\left[\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\left(1+2\sqrt{x}+x\right)\left(1-2\sqrt{x}+x\right)\)
\(=\left(1+\sqrt{x}\right)^2\left(1-\sqrt{x}\right)^2\)
\(=\left[\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\right]^2=\left(1-x\right)^2\)
b/ \(P< 7-4\sqrt{3}\Leftrightarrow\left(1-x\right)^2< 7-4\sqrt{3}\)
\(\Rightarrow\left(1-x\right)^2< \left(2-\sqrt{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-x< 2-\sqrt{3}\Rightarrow x>-1+\sqrt{3}\\1-x< \sqrt{3}-2\Rightarrow x>3-\sqrt{3}\end{cases}}\)
Vậy \(x>3-\sqrt{3}\)
\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
b)Để \(P< \frac{15}{4}\)thì \(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
Ta có:\(\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}< \frac{15}{4}\)
\(\Leftrightarrow\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}-\frac{15}{4}< 0\)
\(\Leftrightarrow\frac{12\sqrt{x}+32-15\sqrt{x}-30}{4\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{-\left(3\sqrt{x}+2\right)}{4\sqrt{x}+8}< 0\)
Vì \(x\ge0;x\ne1\)
Do đó \(0< 4\sqrt{x}+8\)
Mà \(-\left(3\sqrt{x}+2\right)< 0\)
Vậy \(P< \frac{15}{4}\left(đpcm\right)\)
c)Ta có:\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\sqrt{x}+6+2}{\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow P=\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}+\frac{2}{2\sqrt{x}+2}\)
\(\Leftrightarrow P=3+\frac{2}{\sqrt{x}+2}\)
Vì \(x\ge0;x\ne1\Rightarrow\frac{2}{\sqrt{x}+2}\le1\)
Do đó \(P\le4\Leftrightarrow x=1\)
Vậy Max P=4 khi x=1
P=3x+3√x−9(√x−1)(√x+2) +√x+3√x+2 −√x−2√x−1
P=3x+3√x−9(√x−1)(√x+2) +(√x+3)(√x−1)(√x+2)(√x−1) −x−4(√x−1)(√x+2)
P=3x+3√x−9+x+2√x−3−x+4(√x−1)(√x+2)
P=3x−8+5√x(√x−1)(√x+2)
P=3x−3√x+8√x−8(√x−1)(√x+2)
P=(3√x+8)(√x−1)(√x−1)(√x+2)
P=(3√x+8)(√x+2)
b)Để P<154 thì (3√x+8)(√x+2) <154
Ta có:(3√x+8)(√x+2) <154
⇔(3√x+8)(√x+2) −154 <0
⇔12√x+32−15√x−304(√x+2) <0
⇔−(3√x+2)4√x+8 <0
Vì x≥0;x≠1
Do đó 0<4√x+8
Mà −(3√x+2)<0
Vậy P<154 (đpcm)
c)Ta có:P=(3√x+8)(√x+2)
⇔P=3√x+6+2(√x+2)
⇔P=3(√x+2)(√x+2) +22√x+2
⇔P=3+2√x+2
Vì x≥0;x≠1⇒2√x+2 ≤1
Do đó