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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) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a/ Điều kiện xác định tự tìm nhé.
\(\sqrt{a}=\sqrt{2\sqrt{2}+3}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
Vậy \(A=\frac{2\sqrt{2}+3-1}{\sqrt{2}+1}=\frac{2\sqrt{2}+2}{\sqrt{2}+1}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)
b/ \(\frac{a-1}{\sqrt{a}}=a-2\Leftrightarrow a-1=a\sqrt{a}-2\sqrt{a}\)
Đặt \(t=\sqrt{a},t>0\) thì : \(t^2-1=t^3-2t\Leftrightarrow t^3-t^2-2t+1=0\)
Giải pt trên để tìm a.
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
đk: \(a\ge0;a\ne1\)
Ta có:
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-2\left(a^2+2\right)\left(1+\sqrt{a}\right)}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{2a+2\sqrt{a}+2-2a^2\sqrt{a}-2a^2-4-4\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-2a^2\sqrt{a}-2a^2+2a-2\sqrt{a}-2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-a^2\sqrt{a}-a^2+a-\sqrt{a}-1}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
Tại \(a=\sqrt{2}\) thì giá trị của B là:
\(B=\frac{-\left(\sqrt{2}\right)^2.\left(\sqrt{\sqrt{2}}\right)-\left(\sqrt{2}\right)^2+\sqrt{2}-\sqrt{\sqrt{2}}-1}{\left(1+\sqrt{\sqrt{2}}\right)\left(1-\sqrt{\sqrt{2}}\right)\left(\sqrt{2}+\sqrt{\sqrt{2}}+1\right)}\)
\(B\approx3,45267\)
\(ĐKXĐ:x>1\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(=\frac{1-\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{2}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}=\frac{1}{1-a}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-a^2-a-1+a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-1}{a^2+a+1}\)
Với \(a=\sqrt{2}\)( thỏa mãn ĐKXĐ ), ta có:
\(B=\frac{-1}{\left(\sqrt{2}\right)^2+\sqrt{2}+1}=\frac{-1}{2+\sqrt{2}+1}=\frac{-1}{3+\sqrt{2}}\)