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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!
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
a) Với \(x\ne\pm1\)thì \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\right)=\left(\frac{x^2+2x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}\right):\left(\frac{2}{x^2-1}-\frac{x^2+x}{x^2-1}+\frac{x-1}{x^2-1}\right)=\frac{4x}{x^2-1}:\frac{1-x^2}{x^2-1}=\frac{-4x}{x^2-1}\)b) \(x=\sqrt{3+\sqrt{8}}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
Khi đó \(A=\frac{-4\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)^2-1}=\frac{-4\left(\sqrt{2}+1\right)}{2\left(\sqrt{2}+1\right)}=-2\)
c) \(A=\sqrt{5}\Leftrightarrow\frac{-4x}{x^2-1}=\sqrt{5}\Leftrightarrow\sqrt{5}x^2+4x-\sqrt{5}=0\)
Dùng công thức nghiệm của phương trình bậc hai tìm được \(x=\frac{\sqrt{5}}{5}\)hoặc \(x=-\sqrt{5}\)
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
Bài 1 :
a, \(x=25\Rightarrow\sqrt{x}=5\)
Thay vào biểu thức A ta được :
\(A=\frac{25+2.5}{25-1}=\frac{35}{24}\)
b, Với \(x>0;x\ne1\)
\(B=\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+x}{x\left(\sqrt{x}+1\right)}=\frac{2+\sqrt{x}}{x+\sqrt{x}}\)vậy ko xảy ra đpcm
c, Ta có : \(\frac{A}{B}>1\Leftrightarrow\frac{\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-1}}{\frac{2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}}>1\Leftrightarrow\frac{x\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(2+\sqrt{x}\right)}>1\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)do \(\left(\sqrt{x}-1\right)^2+\sqrt{x}\ge0\)
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