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Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
1: Ta có: \(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}\)
2) Ta có: \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=-\left(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(ĐK:x>0;x\ne1\\ A=\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\\ A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}=\dfrac{\left(1-\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(A=\left(\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
nãy đăng ảnh nhưng không hiện, lại phải mất công đánh lại :Đ
a: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b: Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
a) Đk: \(x>0;x\ne9;x\ne25\)
Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)
\(=\dfrac{x}{\sqrt{x}-5}\)
b) Đk: \(x\ge0;x\ne1;x\ne25\)
Biểu thức
\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)
\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x\sqrt{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dkxd:x\ne1,x\ge0\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+2\sqrt{x}+1-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3}{x-1}\)
1: \(=\left(1+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)
2: \(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{1-x}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}+1+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)