\(B=\dfrac{2-x}{2\sqrt{x+x}}-\dfrac{1}{\sqrt{x}}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}\left(x>0;x\ne4\right)\)
a. Tìm số tự nhiên x để B đạt min
b. Tìm x để \(\sqrt{B}>\dfrac{1}{2}\)
Những câu hỏi liên quan
Rút gọn:
A=\(\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right)\div\dfrac{1-\sqrt{x}}{2-\sqrt{x}}vớix>0,x\ne1\)
B=\(\left(\dfrac{x}{3+\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right)\div\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Lm nhanh giúp mk nhé!
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
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12 tháng 4 2021 lúc 0:04
\(B=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)\(\left(x>0,x\ne1\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\left[\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\left(1-x\right)^2}{\left(2\sqrt{x}\right)^2}\)
\(=\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
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Ta có: \(B=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}{4x}\)
\(=\dfrac{-4\sqrt{x}\cdot\left(x-1\right)}{4x}\)
\(=\dfrac{-\left(x-1\right)}{\sqrt{x}}=\dfrac{1-x}{\sqrt{x}}\)
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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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Bài 1 : Cho Adfrac{x}{sqrt{x}-1}
Bleft(dfrac{sqrt{x}}{sqrt{x}-1}+dfrac{sqrt{x}}{x-1}right)divleft(dfrac{2}{x}+dfrac{x+2}{xleft(sqrt{x}-1right)}right)ĐKXĐ : x 0 ; x ≠ 1Tìm GTNN của sqrt{A}Bài 2 :Cho Adfrac{sqrt{x}-2}{3}
Bdfrac{3x+4}{x-2sqrt{x}}+dfrac{2}{sqrt{x}}-dfrac{2sqrt{x}}{sqrt{x}-2}Cho x ∈ N , tìm GTLN của sqrt{B}
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Bài 1 :
Cho \(A=\dfrac{x}{\sqrt{x}-1}\\ B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\div\left(\dfrac{2}{x}+\dfrac{x+2}{x\left(\sqrt{x}-1\right)}\right)\)
ĐKXĐ : x > 0 ; x ≠ 1
Tìm GTNN của \(\sqrt{A}\)
Bài 2 :
Cho \(A=\dfrac{\sqrt{x}-2}{3}\\ B=\dfrac{3x+4}{x-2\sqrt{x}}+\dfrac{2}{\sqrt{x}}-\dfrac{2\sqrt{x}}{\sqrt{x}-2}\)
Cho x ∈ N , tìm GTLN của \(\sqrt{B}\)
\(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x};\left(x\ge0;x\ne9;x\ne16\right)\)
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1};\left(x>0;x\ne1\right)\)
1.
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-15\sqrt{x}}{x-9}\)
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2.
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}+9+2\sqrt{x}-6+x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x}{x-9}\)
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3.
\(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
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Xem thêm câu trả lời
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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\(choP=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)..a,tìmdkxd..b,rút.gọn.P..c,tính.P.khi.x=4-2\sqrt{3}..d,tìm.x.để.P>0..e,tìm.x.để.P< \dfrac{1}{2}\)
b: Ta có: \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
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6 tháng 8 2021 lúc 16:56
Rút gọn biểu thức dạng chữ:Qleft(dfrac{sqrt{x}+2}{x+2sqrt{x}+1}-dfrac{sqrt{x}-2}{x-1}right).left(x+sqrt{x}right) với x ≥0, x ≠1A Aleft(dfrac{1}{sqrt{x}-2}-dfrac{1}{sqrt{x}+2}+dfrac{4sqrt{x}}{4-x}right):dfrac{sqrt{x}+1}{x-4} với x ≥0, x ≠ 4Aleft(dfrac{sqrt{x}}{sqrt{x}+3}+dfrac{2sqrt{x}}{sqrt{x}-3}-dfrac{3x+9}{x-9}right):dfrac{1}{x+6sqrt{x}+9} với x ≥ 0, x ≠ 9Hộ vs ạ
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Rút gọn biểu thức dạng chữ:
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(x+\sqrt{x}\right)\) với x ≥0, x ≠1
A= \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+1}{x-4}\) với x ≥0, x ≠ 4
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\) với x ≥ 0, x ≠ 9
Hộ vs ạ
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
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2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
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3.
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)
\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)
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Rút gọn các biểu thức sau:
a. A = \(\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b. B = \(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) (x > 0 ; x ≠ 1)
\(a,A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2^2-\sqrt{3}^2}\)
\(=\dfrac{4}{1}=4\)
Vậy \(A=4\)
\(b,B=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)
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Rút gọn các biểu thức sau:
a. \(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b. \(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) (x > 0 ; x ≠ 1)
a: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
b: \(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
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