Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Viết như trên rất khó nhìn.
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Viết như trên rất khó nhìn.
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 :<
Bài 1: Rút gọn:
a) \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\) với x>1
b) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)với x>1
c) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\) với x>1
d) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)với x ≠ 4, x ≠ 16,x >0
Mng giúp mk nha
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
Ch.minh
a) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) = \(\dfrac{3\sqrt{x}-1}{x-1}\)
b) (\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)-\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{1}{x-4}\) = -4\(\sqrt{x}\)
Cho A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
B=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
Chứng minh A+B= \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
Help
Cho biểu thức \(A=\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
Tính A khi \(x=\frac{\sqrt{3}+1}{2\sqrt{2}}\)
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!!!
Phân tích thành nhân tử
\(x+\sqrt{x}\)
\(x-\sqrt{x}\)
\(a+3\sqrt{a}-10\)
\(x\sqrt{x}+\sqrt{x}-x-1\)
\(x+\sqrt{x}-2\)
\(x-5\sqrt{x}+6\)
\(x\sqrt{x}-1\)
\(x\sqrt{x}-x+\sqrt{x}-1\)
\(x+2\sqrt{x}-15\)
\(x-2\sqrt{x}-3\)
\(a+\sqrt{a}-6\)
\(x-16\)
\(x+2\sqrt{x}+1\)
\(x-1\)
\(x-2\sqrt{x}+1\)
\(a\sqrt{a}+1\)
\(a+\sqrt{a}-2\)
\(2x-5\sqrt{x}+3\)
\(x-9\)
\(x+\sqrt{x}-6\)
\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}+\frac{1}{2-\sqrt{x}}\)