Cho A = \(\frac{3\sqrt{x}-3}{x\sqrt{x}-2x+2\sqrt{x}-1}-\frac{4x\sqrt{x}-4}{x^3-1}\)(x>1). Rut gon A va tim x de A=1
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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
\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)
=> \(x_1=9-4\sqrt{5}\)
\(x_2=9+4\sqrt{5}\)
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 \(x>3-\sqrt{3}\)
2 tháng 6 2017
\(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x+3}}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-8+5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)}\)
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
2 tháng 6 2017
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
2T
17 tháng 8 2019
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
24 tháng 11 2019
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
24 tháng 11 2019
a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2
\)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:
A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
19 tháng 8 2023
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)