RÚT GỌN BIỂU THỨC:
9) \(A = \left(\dfrac{1}{1 - \sqrt{x}}+\dfrac{1}{1 + \sqrt{x}}\right) : \left(\dfrac{1}{1 - \sqrt{x}}-\dfrac{1}{1 + \sqrt{x}}\right) + \dfrac{1}{1 - \sqrt{x}} \)
Những câu hỏi liên quan
5 tháng 9 2023 lúc 19:47
RÚT GỌN CÁC BIỂU THỨC SAU:21) A left(dfrac{xsqrt{x} + 1}{x - 1} - dfrac{x - 1}{sqrt{x} - 1}right) : left(sqrt{x} + dfrac{sqrt{x}}{sqrt{x} - 1}right) 22) A left(dfrac{x}{sqrt{x} - 1} - sqrt{x}right) : left(dfrac{sqrt{x} + 1}{sqrt{x}} - dfrac{1}{1 - sqrt{x}} + dfrac{2 - x}{x - sqrt{x}}right)23) A left(dfrac{sqrt{x} - 4}{x - 2sqrt{x}} - dfrac{3}{2 - sqrt{x}}right) : left(dfrac{sqrt{x} + 2}{sqrt{x}} - dfrac{sqrt{x}}{sqrt{x} - 2}right)24) A left(dfrac{2x + 1}{xsqrt{x} - 1} + dfrac{1}{1 - sqrt{x}}...
Đọc tiếp
RÚT GỌN CÁC BIỂU THỨC SAU:
21) \(A = \left(\dfrac{x\sqrt{x} + 1}{x - 1} - \dfrac{x - 1}{\sqrt{x} - 1}\right) : \left(\sqrt{x} + \dfrac{\sqrt{x}}{\sqrt{x} - 1}\right) \)
22) \(A = \left(\dfrac{x}{\sqrt{x} - 1} - \sqrt{x}\right) : \left(\dfrac{\sqrt{x} + 1}{\sqrt{x}} - \dfrac{1}{1 - \sqrt{x}} + \dfrac{2 - x}{x - \sqrt{x}}\right)\)
23) \(A = \left(\dfrac{\sqrt{x} - 4}{x - 2\sqrt{x}} - \dfrac{3}{2 - \sqrt{x}}\right) : \left(\dfrac{\sqrt{x} + 2}{\sqrt{x}} - \dfrac{\sqrt{x}}{\sqrt{x} - 2}\right)\)
24) \(A = \left(\dfrac{2x + 1}{x\sqrt{x} - 1} + \dfrac{1}{1 - \sqrt{x}}\right) : \left(1 - \dfrac{x - 2}{x + \sqrt{x} + 1}\right)\)
25) \(A = 1 : \left(\dfrac{x + 2\sqrt{x} - 2}{x\sqrt{x} + 1} - \dfrac{\sqrt{x} -1}{x - \sqrt{x} + 1} + \dfrac{1}{\sqrt{x} + 1}\right)\)
26) \(A = \left(\dfrac{\sqrt{x}}{\sqrt{x} + 2} - \dfrac{3}{2 - \sqrt{x}} + \dfrac{3\sqrt{x} - 2}{x - 2}\right) : \left(\dfrac{\sqrt{x} + 3}{\sqrt{x} - 2} + \dfrac{2\sqrt{x}}{2\sqrt{x} - x}\right)\)
27) \(P = \left(\dfrac{4\sqrt{x}}{2 + \sqrt{x}} + \dfrac{8}{4 - x}\right) : \left(\dfrac{\sqrt{x} - 1}{x - 2\sqrt{x}} - \dfrac{2}{\sqrt{x}}\right)\)
21: ĐKXĐ: x>0; x<>1
\(A=\left(\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{x}\)
\(=\dfrac{-x+\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{1}{x}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}+2}{x}\)
22:
DKXĐ: x>0; x<>1
\(A=\dfrac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1+\sqrt{x}+2-x}\)
\(=\dfrac{x}{\sqrt{x}+1}\)
23: ĐKXĐ: x>0; x<>4
\(A=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)
\(=\dfrac{-4\sqrt{x}+4}{4}=-\sqrt{x}+1\)
24: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
25:
ĐKXĐ: x>=0; x<>1
\(A=1:\dfrac{x+2\sqrt{x}-2-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{2x+\sqrt{x}-1-x+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x+\sqrt{x}}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
27: ĐKXĐ: x>0; x<>4
\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x-8\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-1-2\sqrt{x}+1}\)
\(=\dfrac{4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}}\)
\(=\dfrac{-4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\)
Đúng 1
Bình luận (3)
Rút gọn các biểu thức sau:
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(F=\left(\dfrac{3}{\sqrt{1}+x}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)
\(=\dfrac{3\sqrt{x}}{x-3}\)
f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)
Đúng 2
Bình luận (0)
5 tháng 9 2023 lúc 10:34
RÚT GỌN BIỂU THỨC:
19) \(A = \left(\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x} - 1}\right) : \left(\dfrac{\sqrt{x} + 2}{\sqrt{x} - 1} - \dfrac{\sqrt{x} + 1}{\sqrt{x} - 2}\right)\)
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\) (ĐK: \(x>0;x\ne2;x\ne1\))
\(A=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{-3}\)
\(A=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
Đúng 4
Bình luận (0)
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)\(=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
#Urushi☕
Đúng 2
Bình luận (0)
5 tháng 9 2023 lúc 10:53
RÚT GỌN BIỂU THỨC:
20) \(E = \left(\dfrac{x\sqrt{x} - 1}{x - \sqrt{x}} - \dfrac{x\sqrt{x} +1}{x + \sqrt{x}}\right) + \left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x} + 1}{\sqrt{x} - 1} + \dfrac{\sqrt{x} - 1}{\sqrt{x} + 1}\right)\)
\(E=\left(\dfrac{x\sqrt{x}}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x>0\))
\(E=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{x-\sqrt{x}}-\dfrac{\left(\sqrt{x}\right)^3+1^3}{x+\sqrt{x}}\right]+\left[\dfrac{x}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right]\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(E=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(E=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right)+\dfrac{\left(\sqrt{x}\right)^2-1^2}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{2x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2x+2}{\sqrt{x}}\)
\(E=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
Đúng 4
Bình luận (1)
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!!
Đọc tiếp
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!!
4 tháng 9 2023 lúc 14:30
RÚT GỌN BIỂU THỨC:
12) \(A = \left(\dfrac{x}{\sqrt{x} - 1} - \sqrt{x}\right) : \left(\dfrac{\sqrt{x} + 1}{\sqrt{x}} - \dfrac{1}{1 - \sqrt{x}} + \dfrac{2 - x}{x - \sqrt{x}}\right)\)
\(A=\dfrac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1+\sqrt{x}+2-x}\)
\(=\dfrac{x}{\sqrt{x}+1}\)
Đúng 1
Bình luận (0)
\(A=\left(\dfrac{x}{\sqrt{x}-1}-\sqrt{x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\) (ĐK: \(x>0;x\ne1\))
\(A=\left[\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{x-\sqrt{x}}\right)\)
\(A=\left(\dfrac{x-x+\sqrt{x}}{\sqrt{x}-1}\right):\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}+1}\)
\(A=\dfrac{x}{\sqrt{x}+1}\)
Đúng 2
Bình luận (1)
Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{x}-3}+dfrac{1}{sqrt{x}+3}right)left(1-dfrac{3}{sqrt{x}}right)Bleft(dfrac{sqrt{x}}{sqrt{x}-2}+dfrac{1}{sqrt{x}+2}+dfrac{6-7sqrt{x}}{x-4}right)left(sqrt{x}+2right)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)giúp mình với ạ!mình đang cần gấp
Đọc tiếp
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
Đúng 1
Bình luận (0)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
Đúng 1
Bình luận (0)
3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\times\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
rút gọn biểu thức
Rút gọn biểu thức A = \(\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right):\left(\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{x+1}}\right)\) với \(x=\dfrac{a^2+b^2}{2ab}\)