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`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)
`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)
`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`
`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`
`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`
`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`
`=(x-1)/sqrtx`
`c)A>0`
Mà `sqrtx>0AAx>0`
`<=>x-1>0<=>x>1`
a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)
c, Ta có : \(A>0\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
Vậy ...
\(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a-\sqrt{a}\right)\left(a\sqrt{a}+1\right)}{\left(a-\sqrt{a}\right)\left(a+\sqrt{a}\right)}\)
\(=\frac{a^2\cdot\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}+a-a^2-\sqrt{a}\right)}{a^2-a}\)
\(=\frac{2a^2-2a}{a^2-a}\)
\(=2\)( 1 )
\(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\left(\frac{\sqrt{a}}{1}-\frac{1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\right)\)
\(=\frac{a-1}{\sqrt{a}}\cdot\frac{2\left(a+1\right)}{a-1}\)
\(=\frac{2\left(a+1\right)}{\sqrt{a}}\) ( 2 )
Cộng ( 1 ) và ( 2 ) lại thì ta được biểu thức ban đầu:
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}\)
Câu b,c em chịu:((
P/S:e ko bt đúng hay sai đâu ạ
Mk giải nốt phần còn lại nha
sai thì thông cảm
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}=7\Leftrightarrow2a+2=5\sqrt{a}\)
\(\Leftrightarrow2a-5\sqrt{a}+2=0\)
\(\Leftrightarrow\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{4}\\a=4\end{cases}}\)
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}>6\)\(\Rightarrow2a+2>4\sqrt{a}\Rightarrow2\left(a+1-2\sqrt{a}\right)>0\)
\(\Leftrightarrow\left(a+1-2\sqrt{a}\right)>0\Leftrightarrow\left(\sqrt{a}-1\right)^2>0\)
\(\Leftrightarrow a\ne1;a\ge0\)
\(a,ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
\(b,A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\frac{\sqrt{a}.\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}-1-\sqrt{a}-1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{2\sqrt{a}}\)
\(=\frac{\sqrt{a}.\left(-2\right).2\sqrt{a}}{2\sqrt{a}}\)
\(=-2\sqrt{a}\)
\(c,\)Để A= -4 thì
\(-2\sqrt{a}=-4\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\)
Kết bạn với mình nha ....
a
\(ĐKXĐ:a\ne0;a\ne1;a\ne\sqrt{2}\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{1}\)
\(Q=\frac{\sqrt{a}-2}{\sqrt{a}}\)
b
\(Q>0\Leftrightarrow\frac{\sqrt{a}-2}{\sqrt{a}}>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>\sqrt{2}\)
Điều kiện: \(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}\)
Ta có:
\(1P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)\)
\(=\frac{\sqrt{a}-2}{\sqrt{a}}\)
Bài làm:
a) đkxđ: \(a\ne1;a>0\)
b) Ta có:
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right)\div\frac{\sqrt{a}+1}{a}\)
\(A=\left[\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right].\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}\)
\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(A=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-1\right)\sqrt{a}\)
\(A=a-\sqrt{a}\)
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)