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1, a, ĐKXĐ: x > 0
\(\Rightarrow P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\)
\(\Rightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\)
\(\Rightarrow P=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}\)
\(\Rightarrow P=x+\sqrt{x}-2\sqrt{x}\)
\(\Rightarrow P=x-\sqrt{x}\)
b, Thay x=100 vào biểu thức P, ta có:
P= 100 - \(\sqrt{100}\)
\(\Rightarrow P=100-10=90\)
Vậy với x=100 thì P=90
c, Ta có: P= \(x-\sqrt{x}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi...
2, a, ĐKXĐ: x \(\ge\) 0, x \(\ne\) 1
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\frac{x-1}{1}\)
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1-\sqrt{x}-2-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\frac{x-1}{1}\)
\(\Rightarrow\)A= \(\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{x-1}{1}\)= x-1
b, Để \(\frac{1}{A}\)là số tự nhiên (x \(\ge0\), \(x\ne1\))
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
Vậy x=2 thì \(\frac{1}{A}\) là số tự nhiên.
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
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
mk làm luôn.
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(\frac{3.\left(x+\sqrt{x}\right).\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^