Các câu hỏi tương tự
7 tháng 7 2023 lúc 13:59
Cho 0<x<2. Chứng minh rằng:
\(\dfrac{4-\sqrt{4-x^2}}{\sqrt{\left(2+x\right)^3}+\sqrt{\left(2-x\right)^3}}\) + \(\dfrac{4+\sqrt{4-x^2}}{\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}}\) = \(\dfrac{\sqrt{2+x}}{x}\)
Rút gọn biểu thức sau:
\(\sqrt{12+6\sqrt{3}}-\sqrt{3}\)
\(\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\times\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\)
\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) RÚT GỌN
5 tháng 9 2023 lúc 20:27
RÚT GỌN BIỂU THỨC SAU:
26) \(A = \left(\dfrac{\sqrt{x}}{\sqrt{x} + 2} - \dfrac{3}{2 - \sqrt{x}} + \dfrac{3\sqrt{x} - 2}{x - 2}\right) : \left(\dfrac{\sqrt{x} + 3}{\sqrt{x} - 2} + \dfrac{2\sqrt{x}}{2\sqrt{x} - x}\right)\)
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
29 tháng 3 2021 lúc 23:01
cho biểu thức P=\(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)\(\div\)\(\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
rút gọn P
tìm giá trị để P>0
1.Aleft(dfrac{x+2}{xsqrt{x}-1}+dfrac{sqrt{x}}{x+sqrt{x}+1}+dfrac{1}{1-sqrt{x}}right):dfrac{sqrt{x}}{2}2.Bleft(dfrac{2sqrt{x+x+1}}{sqrt{x}+1}right)left(1-dfrac{x-sqrt{x}}{sqrt{x}-1}right):left(1-sqrt{x}right)3.Cleft(dfrac{sqrt{x}+1}{sqrt{x}-1}-dfrac{sqrt{x}-1}{sqrt{x}+1}dfrac{8sqrt{x}}{x-1}right):left(dfrac{sqrt{x}-x-3}{x-1}-dfrac{1}{sqrt{x}-1}right)Làm chi tiết hộ mình với ak mình đang cần gấp!!!
Đọc tiếp
1.A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
2.B=\(\left(\dfrac{2\sqrt{x+x+1}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
3.C=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
Làm chi tiết hộ mình với ak mình đang cần gấp!!!
Cho biểu thức A = \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\):\(\dfrac{\sqrt{x}-1}{2}\) (\(x\ge0\); \(x\ne1\)). Chứng minh rằng \(A>0\)
\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt[]{x}}{x+\sqrt{x}}\right)\)
\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\)
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
RÚT GON