P = ( \(\dfrac{\sqrt{1}}{x-\sqrt{x}}\)-\(\dfrac{1}{\sqrt{x}-1}\)) (x > 0 ; x\(\ne\)1)
a) Rút Gọn P
b) Tính P Khi x = 9
c) tìm x để P < \(\sqrt{P}\)
mn giúp mình với
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RG: B = [\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\) - \(\dfrac{\sqrt{x}}{1-x}\) ] : [ \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\) ] ; ĐKXĐ: x ≥ 0
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
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\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(B=\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{x+2\sqrt{x}+\sqrt{x}+1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{2x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(B=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
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a : dfrac{x}{x-4}+dfrac{1}{sqrt{x}-2}+dfrac{1}{sqrt{x}+2}với a ≥ 0 x ≠ 4 b : left(dfrac{1}{sqrt{x}}+dfrac{sqrt{x}}{sqrt{x}+1}right).dfrac{sqrt{x}}{x+sqrt{x}}c : dfrac{sqrt{x}}{sqrt{x}-1}+dfrac{3}{sqrt{x}+1}-dfrac{6sqrt{x}-4}{x-1}d : left[dfrac{a+3sqrt{a}+2}{left(sqrt{a}+2right)left(sqrt{a}-1right)}-dfrac{asqrt{a}}{a-1}right]:left(dfrac{1}{sqrt{a}-1}+dfrac{1}{sqrt{a}+1}right)
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a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
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c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
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P(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x+2}}).dfrac{sqrt{x}+1}{sqrt{x}-1} với x 0 và x≠1 A(dfrac{2}{sqrt{x}-2}+dfrac{3}{2sqrt{x+1}}-dfrac{5sqrt{x}-7}{2x-3sqrt{x}-2}):dfrac{2sqrt{x}+3}{5x-10sqrt{x}} với x 0 và x≠4Adfrac{x+1-2sqrt{x}}{sqrt{x-1}}+dfrac{x+sqrt{x}}{sqrt{x}+1} với x≥0 và x≠1V(dfrac{1}{sqrt{x}+2}+dfrac{1}{sqrt{x}-2})dfrac{sqrt{x}+2}{sqrt{x}} với x0 và x≠4A(dfrac{1}{x-1}+dfrac{3sqrt{x}+5}{xsqrt{x}-x-sqrt{x}+1})(dfrac{left(sqrt{x}+1right)^2}{4sqrt{x}}-1)Pdfrac{15sqrt{x}-11}{x...
Đọc tiếp
P=(\(\dfrac{x-2}{x+2\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x+2}}\)).\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với x >0 và x≠1
A=(\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{3}{2\sqrt{x+1}}\)-\(\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\)):\(\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\) với x > 0 và x≠4
A=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x-1}}\)+\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) với x≥0 và x≠1
V=(\(\dfrac{1}{\sqrt{x}+2}\)+\(\dfrac{1}{\sqrt{x}-2}\))\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) với x>0 và x≠4
A=(\(\dfrac{1}{x-1}\)+\(\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\))(\(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}\)-1)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)+\(\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
MỌI NGƯỜI GIÚP ĐỠ MÌNH RÚT GỌN MẤY BIỂU THỨC NÀY VỚI Ạ . EM XIN CẢM ƠN
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(=\dfrac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
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Aleft(dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}right):dfrac{x-2sqrt{x}+1}{x-1} ( ĐKXĐ x0;x≠4)Pleft(dfrac{3sqrt{x}}{sqrt{x}+2}+dfrac{sqrt{x}}{2-sqrt{x}}+dfrac{8sqrt{x}}{x-4}right):left(2-dfrac{2sqrt{x}+3}{sqrt{x}+2}right) Eleft(1-dfrac{sqrt{x}}{sqrt{x}+1}right):left(dfrac{sqrt{x}+2}{sqrt{x}+3}+dfrac{sqrt{x}-3}{2-sqrt{x}}+dfrac{sqrt{x}-2}{x+sqrt{x}-6}right) (ĐKXĐ x≥0;x≠4)RÚT GỌN GIÚP MÌNH VỚI A
Đọc tiếp
A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) ( ĐKXĐ x>0;x≠4)
P=\(\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{8\sqrt{x}}{x-4}\right):\left(2-\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}\right)\)
E=\(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\) (ĐKXĐ x≥0;x≠4)
RÚT GỌN GIÚP MÌNH VỚI A
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
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20 tháng 12 2022 lúc 20:53
Rút gọn P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{x-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{4\sqrt{x}+1}{x+\sqrt{x}}\right)\) với x>0 ; x≠1
\(=\dfrac{x+\sqrt{x}-3\sqrt{x}+1}{x-1}:\dfrac{x+2\sqrt{x}+1-4\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
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Tìm `x >= 0`
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}>1\)
\(\dfrac{2\sqrt{x}}{\sqrt{x}+1}< \dfrac{1}{3}\)
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)
\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)
Mà: \(5\sqrt{x}+10\ge10>0\forall x\)
\(\Leftrightarrow\sqrt{x}>3\)
\(\Leftrightarrow x>9\)
_________
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)
Mà: \(2\sqrt{x}+6\ge6>0\forall x\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
\(\Leftrightarrow0\le x\le1\)
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chứng minh p=\(\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)(x>0)không phụ thuộc vào x
\(P=\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt[4]{x}\left(\sqrt[4]{x}-1\right)}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{x}}+1\right)^2}}{1+\sqrt{x}}\)
\(=\left(-\sqrt[4]{x}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{1}{\sqrt{x}}+1}{1+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}=0\)
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Bài 1: Cho Aleft(dfrac{x-y}{sqrt{x}-sqrt{y}}+dfrac{sqrt{x^3}-sqrt{y^3}}{y-x}right):dfrac{left(sqrt{x}-sqrt{y}right)^2+sqrt{xy}}{sqrt{x}+sqrt{y}}với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}+left(sqrt{x}-dfrac{1}{sqrt{x}}right).left(dfrac{sqrt{x}+1}{sqrt{x}-1}+dfrac{sqrt{x}-1}{sqrt{x}+1}right)
với x0; x≠1
a) Rút gọn A
b)Tìm x để A6
Đọc tiếp
Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
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A1= \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)với x≠1, x≥ 0
A2= \(\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right]:\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)với x≥0, x≠1 và -1
\(A_1=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A_2=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x-\sqrt{x}+1}{x+1}\\ A_2=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-\sqrt{x}+1}\\ A_2=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
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P=[\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] :\(\dfrac{\sqrt{x}-1}{2}\)
a)Rút gọn biểu thức trên
b)Chứng minh rằng P > 0 với mọi x≥ 0 và x ≠ 1.
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)
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a, Với x ≥ 0, x ≠1
P= [ \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] : \(\dfrac{\sqrt{x}-1}{2}\) =
\(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)]
: \(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\):\(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
P= \(\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}\right)^2+2.\dfrac{1}{2}.\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\)= (\(\sqrt{x}+\dfrac{1}{2}\))2 +\(\dfrac{3}{4}\) >\(0\) ∀ x
=> \(\dfrac{3}{x+\sqrt{x}+1}>0\) ∀ x
=> P > 0 với mọi x ≥ 0 và x ≠ 1
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