Chứng minh :
\(x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4};x>0\)
Từ đó, cho biết biểu thức \(\dfrac{1}{x-\sqrt{x}+1}\) có giá trị lớn nhất là bao nhiêu ?
Giá trị đó đạt được khi \(x\) bằng bao nhiêu ?
K
Khách
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NT
Nguyen Thuy Hoa
27 tháng 5 2017
Đúng(0)
Những câu hỏi liên quan
8 tháng 5 2022
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
\(A-1=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-1=\dfrac{\sqrt{x}-2-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+1}\)
Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\forall x\in D\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)
\(\Rightarrow A-1< 0\Rightarrow A< 1\)
1 tháng 5 2022
1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)
=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)
=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)
=\(\dfrac{-4\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
=-4
2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)
=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)
14 tháng 5 2021
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx-4)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
14 tháng 5 2021
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
AT
22 tháng 6 2021
\(P=\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt[4]{x}\left(\sqrt[4]{x}-1\right)}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{x}}+1\right)^2}}{1+\sqrt{x}}\)
\(=\left(-\sqrt[4]{x}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{1}{\sqrt{x}}+1}{1+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}=0\)
N
19 tháng 11 2021
\(a,VT=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{\left(x+1\right)\left(1-3x\right)}{3x}\right)\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2-6x}{3x}\right)\cdot\dfrac{x}{x-1}=\dfrac{6x}{3x}\cdot\dfrac{x}{x-1}=\dfrac{2}{x-1}=VP\left(x\ne0;x\ne1\right)\)
\(b,VT=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}=VP\left(a\ge0;a\ne1\right)\)
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)