\(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
rút gọn
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Rút gọndfrac{6}{sqrt{5}+1}+sqrt{dfrac{2}{3-sqrt{5}}}-dfrac{10}{sqrt{5}} B1. Với xge0,xne4.ChobiểuthứcAdfrac{x-9}{left(sqrt{x}+3right)left(sqrt{x}-2right)}-dfrac{1}{2-sqrt{x}}-dfrac{sqrt{x}}{sqrt{x}+3}Bdfrac{1}{xsqrt{x}+27}a, tính giá trị biểu thức khi B 1/4b, Rút gọn Ac, Tìm giá trị của x để A1/2d, Với C B : A. Tìm GTLN C
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Rút gọn
\(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
B1. Với \(x\ge0,x\ne4.Chobiểuthức\)
\(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{2-\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{1}{x\sqrt{x}+27}\)
a, tính giá trị biểu thức khi B= 1/4
b, Rút gọn A
c, Tìm giá trị của x để A>1/2
d, Với C= B : A. Tìm GTLN C
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
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Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
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Câu 1: Rút gọn biểu thức: Bleft(dfrac{x}{x+3sqrt{x}}+dfrac{1}{sqrt{x}+3}right):left(1-dfrac{2}{sqrt{2}}+dfrac{6}{x+3sqrt{x}}right) với x 0Câu 2: Rút gọn biểu thức:Pdfrac{xsqrt{2}}{2sqrt{x}+xsqrt{2}}+dfrac{sqrt{2x}-2}{x-2} với x 0; x ne 2Câu 3: Rút gọn biểu thức:Qleft(dfrac{a}{a-2sqrt{a}}+dfrac{a}{sqrt{a}-2}right):dfrac{sqrt{a}+1}{a-4sqrt{a}+4} với a 0; a ne 4
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Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
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Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
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Bài 1.Rút gọn A sqrt{x^2+dfrac{2x^2}{3}} với x0Bài 2.Rút gọn biểu thức left(dfrac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+dfrac{sqrt{30}-sqrt{6}}{sqrt{5}-1}right):dfrac{2}{2sqrt{5}-sqrt{6}}Bài 3.Cho ba biểu thức A asqrt{b} + bsqrt{a};B asqrt{a}-bsqrt{b} ;C a-b.Trong ba biểu thức trên biểu thức bằng biểu thức left(sqrt{a}-sqrt{b}right)left(sqrt{a}+sqrt{b}right) với a,b0Bài 7.Cho B dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{98}+sqrt{99}}+dfrac{1}{sq...
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Bài 1.Rút gọn A = \(\sqrt{x^2+\dfrac{2x^2}{3}}\) với x<0
Bài 2.Rút gọn biểu thức \(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right)\):\(\dfrac{2}{2\sqrt{5}-\sqrt{6}}\)
Bài 3.Cho ba biểu thức A = a\(\sqrt{b}\) + b\(\sqrt{a}\);B = \(a\sqrt{a}-b\sqrt{b}\) ;C = a-b.Trong ba biểu thức trên biểu thức bằng biểu thức \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\) với a,b>0
Bài 7.Cho B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).Giá trị của biểu thức B là
Bài 8.Gọi M là giá trị nhỏ nhất của \(\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\) và N là giá trị lớn nhất của \(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\).Tìm M và N
Giúp mình với!Mình đang cần gấp
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
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Rút gọn biểu thức
\(a.\dfrac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(b.x\sqrt{2x+2}+\left(x+1\right)\sqrt{\dfrac{2}{x+1}}-4\sqrt{\dfrac{x+1}{2}}\)
\(a,=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}=-1-3\sqrt{15}\)
\(b,=x\sqrt{2\left(x+1\right)}+\sqrt{\dfrac{2\left(x+1\right)^2}{x+1}}-\sqrt{\dfrac{16\left(x+1\right)}{2}}\\ =x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\\ =\sqrt{2\left(x+1\right)}\left(x+1-2\right)=\left(x-1\right)\sqrt{2\left(x+1\right)}\)
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a.\(=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{5-3}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{5-3}\)
=\(\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}\)
=\(-1-3\sqrt{15}\)
b.=\(x\sqrt{2\left(x+1\right)}+\left(x+1\right)\sqrt{\dfrac{2\left(x+1\right)}{\left(x+1\right)^2}}-4\sqrt{\dfrac{2\left(x+1\right)}{2^2}}\)
=\(x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\)
=\(\sqrt{2\left(x+1\right)}\left(x+1-2\right)\)
=\(\left(x-1\right)\sqrt{2\left(x+1\right)}\)
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23 tháng 8 2021 lúc 19:43
Rút gọn:1) Pdfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-dfrac{2x+sqrt{x}}{sqrt{x}}+dfrac{2left(x-1right)}{sqrt{x}-1}2)Aleft(dfrac{2sqrt{x}+x}{xsqrt{x}-1}-dfrac{1}{sqrt{x}-1}right):left(1-dfrac{sqrt{x}+2}{x+sqrt{x}+1}right)3) Aleft(dfrac{1}{sqrt{a}-1}-dfrac{1}{sqrt{a}}right):left(dfrac{sqrt{a}+1}{sqrt{a}-2}-dfrac{sqrt{a}+2}{sqrt{a}-1}right) 4) Aleft(dfrac{1}{sqrt{x}+1}-dfrac{2sqrt{x}-2}{xsqrt{x}-sqrt{x}+x-1}right):left(dfrac{1}{sqrt{x}-1}-dfrac{2}{x-1}right)Mng giúp e vs ạ, cần gấp :
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Rút gọn:
1) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
2)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
3) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
4) \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
Mng giúp e vs ạ, cần gấp :<
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
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2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
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15 tháng 8 2021 lúc 15:29
Rút gọn:1) Aleft(dfrac{x-5sqrt{x}}{x-25}-1right):left(dfrac{25-x}{x+2sqrt{x}-15}-dfrac{sqrt{x}+3}{sqrt{x}+5}-dfrac{sqrt{x}-5}{sqrt{x}-3}right)2) Aleft(dfrac{1}{sqrt{a}-1}-dfrac{1}{sqrt{a}}right):left(dfrac{sqrt{a}+1}{sqrt{a}-2}-dfrac{sqrt{a}+2}{sqrt{a}-1}right)3) Aleft(dfrac{sqrt{x}+2}{x+2sqrt{x}+1}-dfrac{sqrt{x}-2}{x-1}right).dfrac{sqrt{x}+1}{sqrt{x}}4) Aleft(dfrac{x+1}{x-1}-dfrac{x-1}{x+1}+dfrac{x^2-4x-1}{x^2-1}right).dfrac{x+2003}{x}5) Aleft(dfrac{5sqrt{x}}{x-4}-dfrac{sqrt{x}}{sqrt{x}-2}+dfra...
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Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
4) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
5) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
6) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp mình với, cần gấp ạ 
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
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1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
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13 tháng 8 2021 lúc 23:36
Rút gọn:1) Aleft(dfrac{x-5sqrt{x}}{x-25}-1right):left(dfrac{25-x}{x+2sqrt{x}-15}-dfrac{sqrt{x}+3}{sqrt{x}+5}+dfrac{sqrt{x}-5}{sqrt{x}-3}right)2) Aleft(dfrac{1}{sqrt{a}-1}-dfrac{1}{sqrt{a}}right):left(dfrac{sqrt{a}+1}{sqrt{a}-2}-dfrac{sqrt{a}+2}{sqrt{a}-1}right)3) Aleft(dfrac{x+1}{x-1}-dfrac{x-1}{x+1}+dfrac{x^2-4x-1}{x^2-1}right).dfrac{x+2003}{x}4) Aleft(dfrac{5sqrt{x}}{x-4}-dfrac{sqrt{x}}{sqrt{x}-2}+dfrac{sqrt{x}}{sqrt{x}+2}right)left(2-sqrt{x}right)5) Aleft(dfrac{sqrt{x}+2}{x+2sqrt{x}+1}-dfrac{...
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Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
4) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
5) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp vs ạ 
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
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2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
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3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
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1) Rút gọn biểu thức
P=\(\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right).\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
Ta có: \(P=\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{x\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)\(=\dfrac{3\sqrt{x}-6+\sqrt{x}+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)^2}\)
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RÚT GỌN P:
P=\(\left(2+\dfrac{\sqrt{X}-1}{2\sqrt{X}-3}\right):\left(\dfrac{6\sqrt{X}+1}{2X-\sqrt{X}-3}+\dfrac{\sqrt{X}}{\sqrt{X}+1}\right)\)
\(P=\left(2+\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2\left(2\sqrt{x}-3\right)+\left(\sqrt{x}-1\right)}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{4\sqrt{x}-6+\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right).\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)\)
\(P=\dfrac{\left(5\sqrt{x}-7\right)\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}\)
\(P=\dfrac{5\sqrt{x}-7}{2\sqrt{x}+1}\)
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