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a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
a: \(K=\dfrac{\sqrt{x}-11-3x-9\sqrt{x}-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x-8\sqrt{x}-11-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x-9\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
b: Để K=1/2 thì \(\dfrac{-5x-9\sqrt{x}-8}{x+2\sqrt{x}-3}=\dfrac{1}{2}\)
=>\(-10x-18\sqrt{x}-16=x+2\sqrt{x}-3\)
=>-11x-20căn x+13=0
=>\(x=\left(\dfrac{-10+9\sqrt{3}}{11}\right)^2\)
a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)
c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>căn x-3<0
=>0<x<9
`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)
\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)
\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
`b,`Ta có :
\(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)
\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)
\(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}\)
\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)
\(M=\dfrac{6+9\sqrt{2}}{2}\)
`c,` Để `M<1` Ta có :
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )
\(\sqrt{x}< 3\)
\(x< 9\)
Đối chiếu ĐKXĐ ta có : `0<x<9`
1:
\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
3: A nguyên
=>-5căn x-15+17 chia hết cho căn x+3
=>căn x+3 thuộc Ư(17)
=>căn x+3=17
=>x=196
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)
b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
Câu 2:
a: ĐKXĐ: x>=0; x<>1
b: \(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\dfrac{2\sqrt{x}}{2}\cdot\left(\sqrt{x}-1\right)=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Thay x=4/25 vào G, ta được:
\(G=-\dfrac{2}{5}\cdot\left(\dfrac{2}{5}-1\right)=\dfrac{-2}{5}\cdot\dfrac{-3}{5}=\dfrac{6}{25}\)
a ,rút gọn P (dkxd x\(\ge0,x\ne0\)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x-1}\right)}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
..............=\(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
a: ĐKXĐ: x>=0; x<>1
b: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
c: Để K=1/2 thì \(\dfrac{-5x+5\sqrt{x}-8}{x+2\sqrt{x}-3}=\dfrac{1}{2}\)
=>\(-10x+10\sqrt{x}-16-x-2\sqrt{x}+3=0\)
=>\(-11x+8\sqrt{x}-13=0\)
hay \(x\in\varnothing\)