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a) Thay x = 25 vào biểu thức A , ta có
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
b) \(B=\frac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{x+1+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Ta có : \(x=25\Rightarrow\sqrt{x}=5\)
Thay vào biểu thức A ta được :
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
Vậy với x = 25 thì A = 3/4
b, Với \(x\ge0;x\ne1\)
\(B=\frac{x-5}{x-1}-\frac{2}{\sqrt{x}+1}+\frac{4}{\sqrt{x}-1}\)
\(=\frac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{x-1}=\frac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{x-1}\)
\(=\frac{x+1+2\sqrt{x}}{x-1}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}\pm1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c, Ta có P = A/B hay \(P=\frac{\sqrt{x}-2}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(\sqrt{P}< \frac{1}{2}\)hay \(\sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+1}}< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}-\frac{1}{4}< 0\Leftrightarrow\frac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow3\sqrt{x}-9>0\)do \(4\left(\sqrt{x}+1\right)>0\)
\(\Leftrightarrow3\sqrt{x}>9\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
\(a,x=7-4\sqrt{3}=4-2.2\sqrt{3}+3\) (Thỏa mãn ĐKXĐ)
\(=\left(2-\sqrt{3}\right)^2\)
\(B=\frac{2}{\sqrt{x}-2}=\frac{2}{\sqrt{\left(2-\sqrt{3}\right)^2}-2}\)
\(=\frac{2}{2-\sqrt{3}-2}=-\frac{2\sqrt{3}}{3}\)
\(b,P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}:\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(P=\frac{4}{3}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{4}{3}\)
\(\Leftrightarrow3\left(\sqrt{x}+2\right)=4\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow3\sqrt{x}+6=4\sqrt{x}+4\)
\(\Leftrightarrow6-4=4\sqrt{x}-3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(ko thỏa mãn ĐKXĐ)
=>pt vo nghiệm
d,\(\left(\sqrt{x}+1\right)P-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\frac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\sqrt{x}+2-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow-4\sqrt{x-1}+28=-6x+10\sqrt{5x}\)
\(\Leftrightarrow x=5\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
a, Thay x = 9 vào biểu thức \(A=\frac{\sqrt{x}-2}{\sqrt{x}-1}\) ta được:
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{\sqrt{3^2}-2}{\sqrt{3^2}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
Vậy với x = 9 thì \(A=\frac{1}{2}\)
\(b,\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne1\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
a, Thay x=9 ta có
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
b,\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{1}{\sqrt{x}+1}\)
\(K=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\left(\sqrt{a}-1\right)\)
\(=\frac{a-1}{\sqrt{a}}\Rightarrow\left\{{}\begin{matrix}m=1\\n=-1\end{matrix}\right.\Rightarrow m^2+n^2=2\)
\(A=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\Rightarrow\left\{{}\begin{matrix}m=0\\n=-2\end{matrix}\right.\Rightarrow m-n=2\)
\(A=\frac{x+2}{x-\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\frac{x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{1}{\sqrt{x}-2}\)
Khi \(x=25\): \(B=\frac{1}{\sqrt{25}-2}=\frac{1}{5-2}=\frac{1}{3}\)
\(P=A\div B=\frac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\div\frac{1}{\sqrt{x}-2}=\frac{4\sqrt{x}+1}{\sqrt{x}+1}\)
\(P^2=P+2\Leftrightarrow P^2-P-2=0\Leftrightarrow\left(P-2\right)\left(P+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}P=2\\P=-1\end{cases}}\)
- \(P=2\): \(\frac{4\sqrt{x}+1}{\sqrt{x}+1}=2\Leftrightarrow4\sqrt{x}+1=2\sqrt{x}+2\Leftrightarrow x=\frac{1}{4}\)(tm)
- \(P=-1\): \(\frac{4\sqrt{x}+1}{\sqrt{x}+1}=-1\Leftrightarrow4\sqrt{x}+1=-\sqrt{x}-1\Leftrightarrow\sqrt{x}=-\frac{2}{5}\)(vô nghiệm)
\( a)A = \dfrac{{a - \sqrt a - 6}}{{4 - a}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{a + 2\sqrt a - 3\sqrt a - 6}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 3}}{{\sqrt a - 2}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 2}}{{\sqrt a - 2}} = - 1 \)
\( b)B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}\\ B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{\sqrt x + 1}} \)