P=\(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
Những câu hỏi liên quan
a) \(\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
b) \(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(\sqrt{a}-\dfrac{1}{a}\right)\)
\(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right].\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)
\(\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{a-1}\cdot\dfrac{2\sqrt{a}}{a-1}\)
\(=\dfrac{2\sqrt{a}\left(a+2\sqrt{a}+1-a\sqrt{a}\right)}{\left(a-1\right)^2}\)
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A left(dfrac{a-1}{sqrt{a}-1}-2right)left(dfrac{a-sqrt{a}}{sqrt{a}-1}+1right) ĐK: (a≥0, a≠1)
B left(dfrac{asqrt{a}-a}{a-sqrt{a}}-dfrac{asqrt{a}+1}{a+sqrt{a}}right):dfrac{a+2}{a-2} ĐK: (a0, a≠0, a≠2)
C left(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{a}{a-1}right):left(sqrt{a}-dfrac{sqrt{a}}{sqrt{a}+1}right) ĐK: (a0, a≠1)
D dfrac{a+sqrt{a}}{sqrt{a}}+dfrac{a+4}{sqrt{a}+2} ĐK: (a0)
E left(sqrt{a}-dfrac{1}{sqrt{a}}right)left(dfrac{sqrt{a}-1}{sqrt{a}}+dfrac{1-sqrt{a}}{a+sqrt{a}}right) ĐK: (a0, a≠1)
Đọc tiếp
A= \(\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)\) ĐK: (a≥0, a≠1)
B= \(\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) ĐK: (a>0, a≠0, a≠2)
C= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)\) ĐK: (a>0, a≠1)
D= \(\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}\) ĐK: (a>0)
E= \(\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)\) ĐK: (a>0, a≠1)
A = \(\left(\dfrac{a-1}{\sqrt{a}-1}-2\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-2\right)\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right)=\left(\sqrt{a}+1-2\right)\left(\sqrt{a}+1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)=a-1\)
\(B=\left(\dfrac{a\sqrt{a}-a}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}=\left(\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}=\left(\dfrac{a}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{\left(\sqrt{a}-1\right)\left(a-2\right)}{\sqrt{a}\left(a+2\right)}\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{a}{a-1}\right):\left(\sqrt{a}-\dfrac{\sqrt{a}}{\sqrt{a}+1}\right)=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\dfrac{a}{a-1}\right):\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)-\sqrt{a}}{\sqrt{a}+1}\right)=\dfrac{\sqrt{a}}{a-1}:\dfrac{a}{\sqrt{a}+1}=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{a}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(D=\dfrac{a+\sqrt{a}}{\sqrt{a}}+\dfrac{a+4}{\sqrt{a}+2}=\sqrt{a}+1+\dfrac{a+4}{\sqrt{a}+2}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{a+2\sqrt{a}+\sqrt{a}+2+a+4}{\sqrt{a}+2}=\dfrac{2a+3\sqrt{a}+6}{\sqrt{a}+2}\)
\(E=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}+\dfrac{1-\sqrt{a}}{a+\sqrt{a}}\right)=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a-1+1-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{\left(\sqrt{a}-1\right)\cdot\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}\cdot\sqrt{a}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
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câu1 : a) A dfrac{sqrt{15}-sqrt{12}}{sqrt{5}-2}-dfrac{1}{2-sqrt{3}}
b) left(dfrac{sqrt{a}-2}{sqrt{a}+2}-dfrac{sqrt{a}+2}{sqrt{a}-2}right).left(sqrt{a}-dfrac{4}{sqrt{a}}right)
Câu 2 :
a) A left(2sqrt{4+sqrt{6-2sqrt{5}}}right).left(sqrt{10}-sqrt{2}right)
b) B left(dfrac{sqrt{a}-1}{sqrt{a}+1}+dfrac{sqrt{a}+1}{sqrt{a}-1}right).left(1-dfrac{2}{a+1}right)^2
Đọc tiếp
câu1 : a) A= \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
Câu 2 :
a) A= \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) B= \(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\dfrac{2}{a+1}\right)^2\)
Câu 2:
a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)
b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)
\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)
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P = \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{-4\sqrt{a}}{a-1}=\dfrac{1-a}{\sqrt{a}}\)
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Ta có: \(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a-1}{4a}\cdot\dfrac{-4\sqrt{a}}{1}\)
\(=\dfrac{-a+1}{\sqrt{a}}\)
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\(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
Ta có: \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{a\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}\left[a\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a\sqrt{a}-a-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-2a+a\sqrt{a}-2\sqrt{a}-1}{2}\)
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\(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)\left(\dfrac{a^2-a\sqrt{a}-a\sqrt{a}-a-a-\sqrt{a}}{a-1}\right)\)
\(=\dfrac{a^2-3a\sqrt{a}-2a}{2\sqrt{a}}\)
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\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
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Rút gọn:
A dfrac{4+sqrt{7}}{3sqrt{2}+sqrt{4+sqrt{7}}}+dfrac{4-sqrt{7}}{3sqrt{2}-sqrt{4-sqrt{7}}}
B dfrac{3sqrt{2}+sqrt{11}}{sqrt{2}+sqrt{6+sqrt{11}}}+dfrac{3sqrt{2}-sqrt{11}}{sqrt{2}-sqrt{6-sqrt{11}}}+18
C dfrac{1}{sqrt{3}+sqrt{5}}+dfrac{1}{sqrt{5}+sqrt{7}}+...+dfrac{1}{sqrt{2n+1}+sqrt{2n+3}}với n thuộc N*
D left(sqrt{3}+1right)left(sqrt{5}-1right)left(sqrt{15}-1right)left(7-2sqrt{3}+sqrt{5}right)
Edfrac{left(4+sqrt{3}right)}{sqrt[]{1}+sqrt{3}}+dfrac{left(8+sqrt{15}right)}{sqrt{3}+sqrt...
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Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1
1left(dfrac{sqrt{a}-2}{sqrt{a}+2}-dfrac{sqrt{a}+2}{sqrt{a}-2}right)cdotleft(sqrt{a}cdotdfrac{4}{sqrt{a}}right) (với a0,a≠4)
2left(dfrac{1}{1-sqrt{a}}-dfrac{1}{1+sqrt{a}}right)cdotleft(1-dfrac{1}{sqrt{a}}right) (với a0,a≠1)
3left(dfrac{sqrt{a}-1}{sqrt{a}+1}+dfrac{sqrt{a}+1}{sqrt{a}-1}right)cdotleft(1-dfrac{2}{a+1}right)^2 (với a0,a≠1)
Rút gọn các biểru thức sau
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1\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}\cdot\dfrac{4}{\sqrt{a}}\right)\) (với a>0,a≠4)
2\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\) (với a>0,a≠1)
3\(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\dfrac{2}{a+1}\right)^2\) (với a>0,a≠1)
Rút gọn các biểru thức sau
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
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2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
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từ giả thiết, ta có dfrac{1}{xy}+dfrac{1}{yz}+dfrac{1}{zx}1
đặt left(dfrac{1}{xy};dfrac{1}{yz};dfrac{1}{zx}right)left(a;b;cright)Rightarrow a+b+c1 left(dfrac{ac}{b};dfrac{ab}{c};dfrac{bc}{a}right)left(dfrac{1}{x^2};dfrac{1}{y^2};dfrac{1}{z^2}right)
ta có VTdfrac{1}{sqrt{1+dfrac{1}{x^2}}}+dfrac{1}{sqrt{1+dfrac{1}{y^2}}}+dfrac{1}{sqrt{1+dfrac{1}{z^1}}}sqrt{dfrac{1}{1+dfrac{ac}{b}}}+sqrt{dfrac{1}{1+dfrac{ab}{c}}}+sqrt{dfrac{1}{1+dfrac{bc}{a}}}
dfrac{1}{sqrt{dfrac{b+ac}{b}}}+dfrac{1}{sqrt{dfrac...
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từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
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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