ĐKXĐ: ...
\(\Leftrightarrow x^2+\left(\frac{x}{x+1}\right)^2-2x.\frac{x}{x+1}+\frac{2x^2}{x+1}=1\)
\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2-\frac{2x^2}{x+1}-1=0\)
\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2-\frac{2x^2}{x+1}-1=0\)
Đặt \(\frac{x^2}{x+1}=t\Rightarrow t^2-2t-1=0\)
Bạn tự giải nốt
Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
Rút gọn các biểu thức sau
a,\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
b,\(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
c,\(C=\left(1-\dfrac{x+3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
d,\(D=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
e,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)
B=\(\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)
Giải phương trình:
a) \(2\sqrt{x}\) + 1 = \(\sqrt{2}\) = 5
b) \(\dfrac{\sqrt{x-1}}{\sqrt{x-2}}\)= \(\dfrac{1}{2}\)
c) \(\dfrac{1}{\sqrt{x-3}}\) = \(\dfrac{2}{\sqrt{x-5}}\)
Tính giá trị của P = \(\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)khi x=\(\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
Rút gọn: B= \(\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{1+2x+x^2}+\left(1+\dfrac{1}{x}\right).\dfrac{2}{\left(1+x\right)^3}\right]:\dfrac{x-1}{x^3}\)
C=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a) \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)
b) \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)
\(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)}\)
A=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
