Cho biểu thức :\(A=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
a. Rút gọn biểu thức trên
b. tìm x để A=4
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5 tháng 6 2017
a. ĐKXĐ: \(x\ne0;1\)
Ta có: \(A=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(x+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x^2+2x\sqrt{x}-2\sqrt{x}-1}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b. \(A=4\Leftrightarrow4=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\sqrt{x}\)
\(\Leftrightarrow x+1+2\sqrt{x}-4\sqrt{x}=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)
Vậy....
Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
ĐKXĐ: \(x>0;x\ne1\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\frac{x+2\sqrt{x}+1}{\sqrt{x}}\)
Để \(A=4\Rightarrow\frac{x+2\sqrt{x}+1}{\sqrt{x}}=4\)
\(\Leftrightarrow x+2\sqrt{x}+1=4\sqrt{x}\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1\Rightarrow x=1\) (ko phù hợp ĐKXĐ)
Vậy ko tồn tại x thỏa mãn