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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 \(x=4\Leftrightarrow M=\frac{4}{5}\)
M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)
=\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)
Thuy Duong Nguyen đánh đề cẩn thận hơn bạn nhé
Lời giải :
a) ĐKXĐ : \(x\ne1\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(\Leftrightarrow\sqrt{x}=\sqrt{2}-1\)
Khi đó \(A=\frac{2-5\left(\sqrt{2}-1\right)}{\sqrt{2}-1+3}\)
\(A=\frac{2-5\sqrt{2}+5}{\sqrt{2}+2}=\frac{7-5\sqrt{2}}{\sqrt{2}+2}\)
c) \(A=\frac{1}{2}\)
\(\Leftrightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow2\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)
\(\Leftrightarrow4-10\sqrt{x}-\sqrt{x}-3=0\)
\(\Leftrightarrow1-11\sqrt{x}=0\)
\(\Leftrightarrow11\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{11}\)
\(\Leftrightarrow x=\frac{1}{121}\)( thỏa )
d) A nguyên \(\Leftrightarrow2-5\sqrt{x}⋮\sqrt{x}+3\)
\(\Leftrightarrow-5\left(\sqrt{x}+3\right)+17⋮\sqrt{x}+3\)
Vì \(-5\left(\sqrt{x}+3\right)⋮\sqrt{x}+3\)
\(\Rightarrow17⋮\sqrt{x}+3\)
\(\Rightarrow\sqrt{x}+3\inƯ\left(17\right)=\left\{17\right\}\)( vì \(\sqrt{x}+3\ge3\))
\(\Leftrightarrow\sqrt{x}=14\)
\(\Leftrightarrow x=196\)( thỏa )
Vậy....
\(a,ĐKXĐ:\orbr{\begin{cases}x+2\sqrt{x}+3\ne0\\\sqrt{x}+3\ne0\end{cases}}\)
\(\Leftrightarrow\orbr{ }\sqrt{x}\ne-3\)
Rút gọn: p/s: sau phân số thứ 2 ở mẫu ko có x à? Bạn chép đề sai?
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)