P=\(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\) (\(x\ge0;x\ne1\))
a) rút gọn P
b) tìm \(x\in Z\)sao cho \(P\in Z\)
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giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
Cho biểu thức Pdfrac{3x+sqrt{9x}-3}{x+sqrt{x}-2}-dfrac{sqrt{x}+1}{sqrt{x}+2}+dfrac{sqrt{x}-2}{1-sqrt{x}};xge0,xne1a) Rút gọn P.b) Tính giá trị của P tại x thỏa mãn left|2x-5right|3c) Tìm các giá trị của x để P 3.d) Tìm các giá trị của x để Pdfrac{1}{2}.e) Tìm các giá trị nguyên của x để P có giá trị nguyên.
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Cho biểu thức \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}};x\ge0,x\ne1\)
a) Rút gọn P.
b) Tính giá trị của P tại x thỏa mãn \(\left|2x-5\right|=3\)
c) Tìm các giá trị của x để P = 3.
d) Tìm các giá trị của x để \(P>\dfrac{1}{2}\).
e) Tìm các giá trị nguyên của x để P có giá trị nguyên.
a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
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Rút gọn các biểu thức sau:
a) \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\), \(x\ge0\)
b) \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\), \(x>0\)
c) \(C=\dfrac{y}{2}+\dfrac{3}{4}\sqrt{1+9y^2-6y}-\dfrac{3}{2}\), \(y\le\dfrac{1}{3}\)
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
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\(\dfrac{3X+\sqrt{9X}-3}{X+\sqrt{X}-2}-\dfrac{\sqrt{X}+1}{\sqrt{X}+2}-\dfrac{\sqrt{X}-2}{\sqrt{X}-1}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
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1\(\sqrt{5+2\sqrt{8}}-\sqrt{5-2\sqrt{8}}\) 2)\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\) 3) \(\dfrac{\sqrt{x^2-5x+6}}{\sqrt{x-2}}\) 4)\(\dfrac{\sqrt{\left(x-4\right)^2}}{x^2-5x+4}\) 5) \(\dfrac{3x+1}{\sqrt{9x^2+6x+1}}\)
Cho biểu thức D = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
với \(x\ne9,x\ge0\)
a) Rút gọn D
b)Tìm x để \(D< \dfrac{-1}{4}\)
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
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a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
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Rút gọn:
\(A=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
Rút gọn:
\(A=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2x+3\sqrt{x}-2-x-4\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
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rút gọn biểu thức sau:
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{3x+8\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{20\sqrt{x}+6x-10}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
Đề sai không vậy?
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Ta có: \(N=\dfrac{3x+5\sqrt{x}}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{3x+5\sqrt{x}+3x+9\sqrt{x}-\sqrt{x}-3+7\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6x+20\sqrt{x}-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
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1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
