Chứng minh rằng: \(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\) với mọi \(x\),\(x\ne0\)
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24 tháng 8 2021
Ta có: \(A-3\)
\(=\dfrac{x+5-3\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}+1}\ge0\forall x\) thỏa mãn ĐKXĐ
hay A\(\ge3\)
T
6 tháng 12 2018
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
3 tháng 8 2021
b) Thay x=49 vào A, ta được:
\(A=\dfrac{7-1}{7-5}=\dfrac{6}{2}=3\)
3 tháng 8 2021
a) Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)
\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
23 tháng 10 2021
a) \(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+1=3\sqrt{x}-3\Leftrightarrow2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
16 tháng 5 2023
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
11 tháng 7 2021
Ta có: \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x-3-2x+8-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Ta có: \(A-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-1\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
hay A>1
Giải:
\(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\)
\(\Leftrightarrow x^4\left(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\right)\ge0\)
\(\Leftrightarrow x^{12}-x^9-x^3+1\ge0\)
\(\Leftrightarrow x^9\left(x^3-1\right)-\left(x^3-1\right)\ge0\)
\(\Leftrightarrow\left(x^3-1\right)\left(x^9-1\right)\ge0\)
\(\Leftrightarrow\left(x^3-1\right)\left(x^3-1\right)\left(x^6+x^3+1\right)\ge0\)
\(\Leftrightarrow\left(x^3-1\right)^2\left(x^6+x^3+1\right)\ge0\) (luôn đúng)
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