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bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
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) ĐKXĐ : \(x\ge0\) và \(x\ne4\)
Rút gọn :\(A=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{4\sqrt{x}}{4-x}\)
\(A=\frac{2\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{4-2\sqrt{x}+2+\sqrt{x}+4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{6+3\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3}{2-\sqrt{x}}\)
b) thay \(x=7+4\sqrt{3}\) vào A
ta được :\(A=\frac{3}{2-\sqrt{7+4\sqrt{3}}}=\frac{3}{2-2+\sqrt{3}}=\frac{3}{\sqrt{3}}\)
vậy vói \(x=7+4\sqrt{3}\) thì \(A=\frac{3}{\sqrt{3}}\)
c)với\(x\ge0\) và \(x\ne4\)
Để \(A=-\frac{3}{7}\Leftrightarrow\frac{3}{2-\sqrt{x}}=-\frac{3}{7}\)
\(\Leftrightarrow3.7=-3\left(2-\sqrt{x}\right)\)
\(\Leftrightarrow21=-6+3\sqrt{x}\)
\(\Leftrightarrow21+6=3\sqrt{x}\)
\(\Leftrightarrow27=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=9\)
\(\Leftrightarrow x=81\)
Vậy để\(A=-\frac{3}{7}\Leftrightarrow x=81\)