Cho P: \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
CMR: P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
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31 tháng 10 2021
a) ĐKXĐ: \(x>0,x\ne1\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}=\dfrac{2\sqrt{x}}{x+\sqrt{x}+1}\)
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
14 tháng 9 2023
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\) (ĐK: \(x\ne4;x>0\))
\(=\left[\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-2\sqrt{x}+\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left[\dfrac{-\left(\sqrt{x}+2\right)^2+\left(\sqrt{x}-2\right)^2-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-x-4\sqrt{x}-4+x+4\sqrt{x}+4-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{-4x}\)
\(=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{-4x}\)
\(=-\dfrac{3\sqrt{x}+6-x-2\sqrt{x}}{4x}\)
\(=-\dfrac{\sqrt{x}-x+6}{4x}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 9 2023
Mình sửa lại phần mẫu số của 3 dòng cuối nhé !
\(-4x\Rightarrow-4x\sqrt{x}\)
\(4x\Rightarrow4x\sqrt{x}\)
17 tháng 8 2021
1: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
2 tháng 6 2018
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)
Tương tự cho 2 BĐT trên ta có:
\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)=\dfrac{1}{3}VT\)
Xảy ra khi \(x=y=z\)
\(P=\left(\dfrac{x-2+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)