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11 tháng 8 2021
nãy đăng ảnh nhưng không hiện, lại phải mất công đánh lại :Đ
11 tháng 8 2021
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: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
23 tháng 8 2021
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
23 tháng 8 2021
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: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
a) \(A=\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(A=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(A=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(A=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2\)
\(A=\left[\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\right]^2\)
\(A=\left(x-1\right)^2\)
\(A=x^2+2x+1\)
27 tháng 10 2021
1) \(A=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\dfrac{2\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{5}{4}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Rightarrow2\sqrt{x}-1< \sqrt{x}+1\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
27 tháng 10 2021
\(1,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ 2,x=9\Leftrightarrow A=\dfrac{6-1}{3+1}=\dfrac{5}{4}\\ 3,A< 1\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)
Y
1 tháng 7 2023
\(a,A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(b,x=14-6\sqrt{5}=\sqrt{5^2}-2.3.\sqrt{5}+3^2=\left(\sqrt{5}-3\right)^2\)
\(\Rightarrow A=\dfrac{\sqrt{\left(\sqrt{5}-3\right)^2}}{14-6\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}+1}\)
\(=\dfrac{\left|\sqrt{5}-3\right|}{-6\sqrt{5}+15+\left|\sqrt{5}-3\right|}\)
\(=\dfrac{3-\sqrt{5}}{-6\sqrt{5}+15+3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{18-7\sqrt{5}}\)
\(c,A=1\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}-x-\sqrt{x}-1=0\Leftrightarrow-x-1=0\Leftrightarrow x=-1\left(ktm\right)\)
Vậy khi A = 1 thì không có giá trị x thỏa mãn.
2 tháng 10 2021
\(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\left(x>0;x\ne1;x\ne\dfrac{1}{4}\right)\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
2 tháng 10 2021
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
D
A= (2/x-√x - 1/√x-1) : x-4/x√x+√x - 2x với x>0, x khác 1, x khác 4 a) rút gọn A b) tìm x để A > -1/2
1
5 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{2}{x-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-4}{x\sqrt{x}+\sqrt{x}-2x}\)
\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{x-4}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\sqrt{x}+1}{\sqrt{x}+2}\)