Cho biểu thức P=(\(\frac{1-x\sqrt{x}}{1-\sqrt{x}}\))+\(\sqrt{x}\)).(\(\frac{1+x\sqrt{x}}{1+\sqrt{x}}\)-\(\sqrt{x}\))
a) Rút gọn P
b) Tìm x để P<7-4\(\sqrt{3}\)
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20 tháng 10 2023
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2:
a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(2P=2\sqrt{x}+5\)
=>\(P=\sqrt{x}+\dfrac{5}{2}\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)
=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)
=>\(2x+3\sqrt{x}-2=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
20 tháng 10 2023
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
AT
11 tháng 6 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Để \(P< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Rightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\sqrt{x}+2}< 0\Rightarrow\dfrac{\sqrt{x}-3}{2\sqrt{x}+2}< 0\)
mà \(2\sqrt{x}+2>0\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
\(\Rightarrow0\le x< 9\left(x\ne1\right)\)
5 tháng 6 2017
a. ĐKXĐ: \(x\ne0;1\)
Ta có: \(A=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(x+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x^2+2x\sqrt{x}-2\sqrt{x}-1}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b. \(A=4\Leftrightarrow4=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\sqrt{x}\)
\(\Leftrightarrow x+1+2\sqrt{x}-4\sqrt{x}=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)
Vậy....
NV
13 tháng 7 2016
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
a) ĐK: \(x\ge0;x\ne1\)
Trước tiên chúng ta tính:
\(1-x\sqrt{x}=1-\left(\sqrt{x}\right)^3=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)
\(1+x\sqrt{x}=1+\left(\sqrt{x}\right)^3=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)\)
khi đó:
P = \(\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)^2\)
\(=\left(x-1\right)^2\)
b) \(P< 7-4\sqrt{3}=4-2.2.\sqrt{3}+3=\left(2-\sqrt{3}\right)^2\)
=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)
<=> \(\sqrt{3}-2< x-1< 2-\sqrt{3}\)
<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)
Đối chiếu điều kiện: \(\sqrt{3}-1< x< 3-\sqrt{3}\) và x khác 1.