Cho các bt: A=\(\frac{\sqrt{x}+1}{\sqrt{x}-4},B=\frac{2\sqrt{x}+1}{x-7\sqrt{x}+12}-\frac{\sqrt{x}+3}{\sqrt{x}-4}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a,tìm x để biểu thức B có nghĩa
b,Rút gọn bt B
c,Tìm x để M=B.A có gt nguyên
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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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
a, B= \(\frac{2\sqrt{x}+1}{x-7\sqrt{x}+12}-\frac{\sqrt{x}+3}{\sqrt{x}-4}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
<=> \(B=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-4}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
Để B có nghĩa
<=> \(\left\{{}\begin{matrix}\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\ne0\\x\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\sqrt{x}\ne4\\\sqrt{x}\ne3\\x\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ne16\\x\ne9\\x\ge0\end{matrix}\right.\)
<=> \(x\ge0,x\ne16,x\ne9\)
Vậy để B có nghĩa <=> \(x\ge0,x\ne16,x\ne9\)
b, Có B=\(\frac{2\sqrt{x}+1}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-4}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)( đk: x\(\ge0\), \(x\ne16,x\ne9\))
<=> \(B=\frac{2\sqrt{x}+1-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}+1-x+9+2x-8\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)=\(\frac{x-5\sqrt{x}+6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\frac{x-2\sqrt{x}-3\sqrt{x}+6}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}-2}{\sqrt{x}-4}\)
ý c, đúng đề chưa bạn