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E = \(x^2-x+\dfrac{1}{2}>0\) với mọi giá trị của x
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HP
14 tháng 8 2021
\(x^2-\sqrt{x}+\dfrac{1}{2}\)
\(=x^2-x+\dfrac{1}{4}+x-\sqrt{x}+\dfrac{1}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\\sqrt{x}-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\) vô nghiệm
Vậy \(x^2-\sqrt{x}+\dfrac{1}{2}>0\forall x\ge0\)
22 tháng 10 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(A=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{x-1}{x-\sqrt{x}}\cdot\left(\sqrt{x}+1\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi \(x=\left(\sqrt{3}-1\right)^2\) thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\left(\sqrt{3}+1\right)}{2}=\dfrac{3\sqrt{3}+3}{2}\)
c: \(P-2=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}-2\)
\(=\dfrac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}>0\)
=>P>2
CM
27 tháng 9 2018
a) Rút gọn E Þ đpcm.
b) Điều kiện xác định E là: x ≠ ± 1
Rút gọn F ta thu được F = 4 Þ đpcm
28 tháng 7 2021
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
29 tháng 7 2021
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
14 tháng 4 2023
a: \(A=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: \(A-5=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>=0\)
=>A>=5
5 tháng 2 2020
a)\(\frac{-1}{4x+2}< 0\)
\(\Leftrightarrow4x+2>0\)
\(\Leftrightarrow4x>-2\)
\(\Leftrightarrow x>\frac{-1}{2}\)
Vậy ...
b)\(\frac{-x^2-2x-3}{x^2+1}\)
Ta có: \(-x^2-2x-3=-\left(x+1\right)^2-2\)
Vì \(-\left(x+1\right)^2\le0;\forall x\)
\(\Rightarrow-\left(x+1\right)^2-2\le-2< 0;\forall x\)
Lại có \(x^2\ge0;\forall x\)
\(\Rightarrow x^2+1\ge1>0;\forall x\)
\(\Rightarrow\frac{-x^2-2x-3}{x^2+1}< 0;\forall x\)
B
2
11 tháng 11 2023
Đặt \(A=\dfrac{x^2+x+1}{-2x^2+2x-2}\)
\(x^2+x+1=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)
\(-2x^2+2x-2\)
\(=-2\left(x^2-x+1\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{2}< =-\dfrac{3}{2}< 0\forall x\)
Do đó: \(A=\dfrac{x^2+x+1}{-2x^2+2x-2}< 0\forall x\)
T
11 tháng 11 2023
\(\dfrac{x^2+x+1}{-2x^2+2x-2}=\dfrac{x^2+x+1}{-2\left(x^2-x+1\right)}\)
Ta thấy:
\(x^2+x+1\\=x^2+2\cdot x\cdot\dfrac12+\left(\dfrac12\right)^2-\left(\dfrac12\right)^2+1\\=\left(x+\dfrac12\right)^2+\dfrac34\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
hay \(x^2+x+1>0\forall x\) (1)
Lại có:
\(x^2-x+1\\=x^2-2\cdot x\cdot\dfrac12+\left(\dfrac12\right)^2-\left(\dfrac12\right)^2+1\\=\left(x-\dfrac12\right)^2+\dfrac34\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{x^2+x+1}{x^2-x+1}>0\forall x\)
\(\Rightarrow\dfrac{x^2+x+1}{-2\left(x^2-x+1\right)}< 0\forall x\)
hay đa thức \(\dfrac{x^2+x+1}{-2x^2+2x-2}< 0\forall x\)
\(\text{#}Toru\)
\(E=x^2-x+\dfrac{1}{2}=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}>0\)
Vậy E > 0 với mọi x