Chứng minh rằng
\(2x+\dfrac{1}{\left(x+1\right)^2}\ge1,\forall x>-1\).
Những câu hỏi liên quan
Chứng minh rằng:
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\left(\forall x,y,z>o\right)\)
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Chứng minh rằng nếu x>-1 thì \(2x+\frac{1}{\left(x+1\right)^2}\ge1\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
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
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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 \(x=-\dfrac{127}{64}\)
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chứng minh rằng :
a, x+2y+dfrac{25}{x}+dfrac{27}{y^2}ge 19 ( forallx,y 0 )
b, x+dfrac{1}{left(x-yright)y}ge3 ( forallxy0 )
c,dfrac{x}{2}+dfrac{16}{x-2}ge13left(forall x2right)
d, a+dfrac{1}{a^2}gedfrac{9}{4}left(forall xge2right)
e, a+dfrac{1}{aleft(a-bright)^2}ge2sqrt{2} ( forall xyge0)
f, dfrac{2a^3+1}{4bleft(a-bright)}ge3[forall agedfrac{1}{2};dfrac{a}{b}1]
g, x+dfrac{4}{left(x-yright)left(y+1right)^2}ge3left(forall xyge0right)
h, 2a^4+dfrac{1}{1+a^2}ge3a^2-1
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chứng minh rằng :
a, x+2y+\(\dfrac{25}{x}\)+\(\dfrac{27}{y^2}\)\(\ge\) 19 ( \(\forall\)x,y \(\)> 0 )
b, \(x+\dfrac{1}{\left(x-y\right)y}\ge3\) ( \(\forall\)x>y>0 )
c,\(\dfrac{x}{2}+\dfrac{16}{x-2}\ge13\left(\forall x>2\right)\)
d, \(a+\dfrac{1}{a^2}\ge\dfrac{9}{4}\left(\forall x\ge2\right)\)
e, a+\(\dfrac{1}{a\left(a-b\right)^2}\ge2\sqrt{2}\) ( \(\forall x>y\ge0\))
f, \(\dfrac{2a^3+1}{4b\left(a-b\right)}\ge3[\forall a\ge\dfrac{1}{2};\dfrac{a}{b}>1]\)
g, x+\(\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge3\left(\forall x>y\ge0\right)\)
h, \(2a^4+\dfrac{1}{1+a^2}\ge3a^2-1\)
Chứng minh rằng:
\(x+\dfrac{1}{x}\ge2\left(\forall x>0\right)\)
ap dung BDT co si cho 2 so ko am
\(x+\dfrac{1}{x}\ge2\sqrt{x.\dfrac{1}{x}}\)
<=>\(x+\dfrac{1}{x}\ge2\) (dpcm)
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Tìm hàm số f(x) thỏa mãn
a)\(f\left(x-1\right)+3f\left(\dfrac{1-x}{1-2x}\right)=1-2x,\forall x\ne\dfrac{1}{2}\)
b)\(f\left(x\right)+f\left(\dfrac{1}{1-x}\right)=x+1-\dfrac{1}{x},\forall x\ne0;x\ne1\)
c) \(3f\left(x\right)-2f\left(f\left(x\right)\right)=x,\forall x\in Z\)
chứng minh rằng :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b)left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3x}-x-1right)right]:dfrac{x+1}{x}dfrac{2x}{x-1}
c)left[dfrac{2}{left(x+1right)^3}left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
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chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
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Cho x,y,z là các số thực khác 1 thoả mãn xyz=1 . Chứng minh rằng
\(\dfrac{x^2}{\left(x-1\right)^2}+\dfrac{y^2}{\left(y-1\right)^2}+\dfrac{z^2}{\left(z-1\right)^2}\ge1\)
Jup mik vs nha mik can gap lam
Đặt \(x=\dfrac{c^2}{ab}\); \(y=\dfrac{a^2}{bc}\); \(z=\dfrac{b^2}{ac}\)
\(\Rightarrow xyz=1\) là điều hiển nhiên
BĐT cần chứng minh tương đương
\(\dfrac{\left(\dfrac{c^2}{ab}\right)^2}{\left(\dfrac{c^2}{ab}-1\right)^2}+\dfrac{\left(\dfrac{a^2}{bc}\right)^2}{\left(\dfrac{a^2}{bc}-1\right)^2}+\dfrac{\left(\dfrac{b^2}{ac}\right)^2}{\left(\dfrac{b^2}{ac}-1\right)^2}\ge1\)
\(\Leftrightarrow\dfrac{c^4}{\left(c^2-ab\right)^2}+\dfrac{a^4}{\left(a^2-bc\right)^2}+\dfrac{b^4}{\left(b^2-ac\right)^2}\ge1\)
Áp dụng BĐT C.B.S
\(\dfrac{c^4}{\left(c^2-ab\right)^2}+\dfrac{a^4}{\left(a^2-bc\right)^2}+\dfrac{b^4}{\left(b^2-ac\right)^2}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(c^2-ab\right)^2+\left(a^2-bc\right)^2+\left(b^2-ac\right)^2}\)ta phải chứng minh:
\(\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(c^2-ab\right)^2+\left(a^2-bc\right)^2+\left(b^2-ac\right)^2}\ge1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\ge a^4+b^4+c^4+a^2b^2+b^2c^2+a^2c^2-2\left(abc^2+a^2bc+b^2ac\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+abc^2+a^2bc\right)\ge0\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2\ge0\) ( luôn đúng )
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Cho dãy số (\(u_n\)) xác định bởi: \(\left\{{}\begin{matrix}0< u_n< 1\\u_n\left(1-u_{n+1}\right)>\dfrac{1}{4},\forall n\ge1\end{matrix}\right.\)
Chứng minh dãy số (\(u_n\)) có giới hạn hữu hạn khi \(n\rightarrow\infty\)
\(u_n-u_{n+1}=u_n+\left(1-u_{n+1}\right)-1\ge2\sqrt{u_n\left(1-u_{n+1}\right)}-1>0\)
\(\Rightarrow u_n>u_{n+1}\Rightarrow\) dãy giảm
Dãy giảm và bị chặn dưới bởi 0 nên có giới hạn hữu hạn.
Gọi giới hạn đó là k
\(\Rightarrow k\left(1-k\right)\ge\dfrac{1}{4}\Rightarrow\left(2k-1\right)^2\le0\Rightarrow k=\dfrac{1}{2}\)
Vậy \(\lim\left(u_n\right)=\dfrac{1}{2}\)
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