Rút gọn:
\(E=\left(1-a^2\right):\left[\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
Những câu hỏi liên quan
Rút gọn:
\(A=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
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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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Câu 1: Rút gọn biểu thức: Aleft(dfrac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(dfrac{1-sqrt{a}}{1-a}right)^2(với a ge 0;a ne1)Câu 2: Rút gọn biểu thức: Mleft(dfrac{a+sqrt{a}}{sqrt{a}+1}+1right)left(1+dfrac{a-sqrt{a}}{1-sqrt{a}}right)(với age0; ane1)
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Câu 1: Rút gọn biểu thức: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)(với a \(\ge\) 0;a \(\ne\)1)
Câu 2: Rút gọn biểu thức: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)(với a\(\ge\)0; a\(\ne\)1)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
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Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
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1. Cho biểu thức: A=\(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
Rút gọn biểu thức trên
A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)
=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)
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Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
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Cho biểu thức \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a. Rút gọn A
b. Tìm \(x\) để \(A>-6\)
c. Tính A khi \(a^2-3=0\)
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
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Rút gọn:
\(A=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(A=\left[1:\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right)\right]\cdot\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\)
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Rút gọn các biểu thức sau:Cleft(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{sqrt{a}}{a-sqrt{1}}right):dfrac{sqrt{a}+1}{a-1}Dleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right).dfrac{sqrt{x}+1}{sqrt{x}-1}Eleft(1+dfrac{x+sqrt{x}}{sqrt{x}+1}right)left(1+dfrac{x-sqrt{x}}{1-sqrt{x}}right)Làm ơn giúp mình với ạ!!mình đang cần gấp lắm!!
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Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
Làm ơn giúp mình với ạ!!mình đang cần gấp lắm!!
rút gọn Q=\(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
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rút gọn
\(\left(\dfrac{\sqrt{a}-2}{a-1}-\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right)\)\(\left(1+\dfrac{1}{\sqrt{a}}\right)\)
ĐK: \(x>0;a\ne1\)
\(\left(\dfrac{\sqrt{a}-2}{a-1}-\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right)\left(1+\dfrac{1}{\sqrt{a}}\right)\)
\(=\left[\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}\right].\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right).\dfrac{1}{\sqrt{a}}\)
\(=\left[\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right].\dfrac{1}{\sqrt{a}}\)
\(=\left[\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{a+\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right].\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}.\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{-2}{a-1}\)
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ĐKXĐ: \(\left\{{}\begin{matrix}a\ne1\\a>0\end{matrix}\right.\)
\(\left(\dfrac{\sqrt{a}-2}{a-1}-\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right)\left(1+\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}+\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}\right).\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}-\dfrac{a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}\right).\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{a-\sqrt{a}-2-a+\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{-4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{-4}{\left(\sqrt{a}-1\right).\sqrt{a}.\left(\sqrt{a}+1\right)}\)
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