Cho biểu thức B=\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}\)- \(\frac{\sqrt{x}-4}{\sqrt{x}+1}\)+ \(\frac{\sqrt{x}+8}{4-\sqrt{x}}\)
a, Rút gọn B
b,TÌm x để B= \(\frac{-4}{\sqrt{x}+3}\)
c, Tìm x để bthức A= \(\frac{2x+1}{\sqrt{x}+2}\).B đạt gtrị nhỏ nhất
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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(đkxđ\Leftrightarrow x\ge4\)
\(P=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)
\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\sqrt{\frac{4^2}{x^2}-2.\frac{4}{x}+1}}\)
\(=\frac{\sqrt{\left(x-4+2\right)^2}+\sqrt{\left(x-4-2\right)^2}}{\sqrt{\left(\frac{4}{x}-1\right)^2}}\)
\(=\frac{|x-2|+|x-6|}{|\frac{4}{x}-1|}=\frac{x-2+|x-6|}{|\frac{4}{x}-1|}\)
Dùng bảng xét dấu nha
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne16\end{cases}}\)
\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)
\(\Leftrightarrow B=\frac{2x+8+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
b) Để B nguyên'
\(\Leftrightarrow3\sqrt{x}⋮\sqrt{x}+1\)
\(\Leftrightarrow3\left(\sqrt{x}+1\right)-3⋮\sqrt{x}+1\)
\(\Leftrightarrow3⋮\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)(Đã loại những giá trị âm)
\(\Leftrightarrow x\in\left\{0;4\right\}\)
Vậy để \(B\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)
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 \(x=4\Leftrightarrow M=\frac{4}{5}\)