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Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)
d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)
\(\Leftrightarrow11\sqrt{x}=1\)
hay x=1/121
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
1/
a) \(\left(\dfrac{2\sqrt{2}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\dfrac{2\sqrt{2}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{2}\cdot\left(\sqrt{x}-3\right)+\sqrt{x}\cdot\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x-3}}\)
\(=\dfrac{2\sqrt{2x}-6\sqrt{2}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{2x}-6\sqrt{2}-2x+3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{2x}-6\sqrt{2}-2x+3\sqrt{x}-3}{x+\sqrt{x}+3\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{2x}-6\sqrt{2}-2x+3\sqrt{x}-3}{x+4\sqrt{x}+3}\)
bài 2 : đk : \(x\ge0;x\ne1\)
a) P = \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
P = \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
P = \(\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\) P = \(\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
P = \(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
P = \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) P = \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(\sqrt{x}+3=4-10\sqrt{x}\)
\(\Leftrightarrow\) \(11\sqrt{x}-1=0\) \(\Leftrightarrow\) \(11\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{11}\) \(x=\left(\dfrac{1}{11}\right)^2=\dfrac{1}{121}\)
a: ĐKXĐ: x>=0; x<>9
\(B=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Thay \(x=11+6\sqrt{2}\) vào B, ta được:
\(B=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}\)
\(=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)
c: Để B<1 thì B-1<0
\(\Leftrightarrow\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>căn x-3<0
=>0<=x<9
a: ĐKXĐ: x>0; x<>9
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+2}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+2}\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
b: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10\sqrt{x}+2=\sqrt{x}+3\)
hay \(x\in\varnothing\)
a: ĐKXĐ: x>0; x<>9
b: \(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để D<-1 thì D+1<0
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}+4< 0\)
\(\Leftrightarrow4-\sqrt{x}< 0\)
hay x>16
a) điều kiện : \(x>0;x\ne9\)
ta có : \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\) \(\Leftrightarrow A=\left(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\right)\) \(\Leftrightarrow A=\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\) b) \(A< -1\) \(\Leftrightarrow A+1< 0\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+3}+1< 0\Leftrightarrow\dfrac{-\sqrt{x}+3}{2\sqrt{x}+3}< 0\)\(\Leftrightarrow-\sqrt{x}+3< 0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
vậy ...
----------------------------(-_-)---------------------------------------------------
A=\(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
=\(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x+3}\right)\left(\sqrt{x}-3\right)}\times\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
=\(\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(a.ĐKXĐ:\left\{{}\begin{matrix}x\ne9\\x\ge0\end{matrix}\right.\)
\(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{x-9}=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
\(b.\dfrac{1}{M}=\dfrac{\sqrt{x}-3}{3\sqrt{x}}\)
Ta có : \(\dfrac{\sqrt{x}-3}{3\sqrt{x}}< \dfrac{1}{6}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-6-\sqrt{x}}{6\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-6}{6\sqrt{x}}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-6< 0\\\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-6>0\\\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 36\\x>0\end{matrix}\right.\) \(\left(x\ne9;x\ne0\right)\)
KL............
P/s : Mk ko quen làm toán 9 lắm , ms lớp 8 thui :)) Nên sai bỏ qua nhe.
cảm ơn nhé!