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Câu 1:
\(A=\dfrac{81x}{3-x}+\dfrac{3}{x}=\dfrac{81x}{3-x}+\left(\dfrac{3}{x}-1\right)+1=\dfrac{81x}{3-x}+\dfrac{3-x}{x}+1\ge2\sqrt{\dfrac{81x}{3-x}.\dfrac{3-x}{x}}+1=18+1=19\)
Dấu "=" xảy ra <=> x = 0,3
Câu 2:
\(\dfrac{1}{3x-2\sqrt{6x}+5}=\dfrac{1}{\left(3x-2\sqrt{6x}+2\right)+3}=\dfrac{1}{\left(x\sqrt{3}-\sqrt{2}\right)^2+3}\le\dfrac{1}{3}\)
Dấu "=" xảy ra <=> \(x=\sqrt{\dfrac{2}{3}}\)
Câu 3:
\(A=2014\sqrt{x}+2015\sqrt{1-x}=2014\left(\sqrt{x}+\sqrt{1-x}\right)+\sqrt{1-x}\)
Ta có: \(\left(\sqrt{x}+\sqrt{1-x}\right)^2=x+1-x+2\sqrt{x\left(1-x\right)}=1+2\sqrt{x\left(1-x\right)}\ge1\)
=> \(A=2014\left(\sqrt{x}-\sqrt{1-x}\right)+\sqrt{1-x}\ge2014+\sqrt{1-x}\ge2014\)
Dấu "=" xảy ra <=> x = 1
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
\(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)
\(S=\frac{a^2+b^2}{ab}\ge\frac{2ab}{ab}=2\)
Dấu \(=\)xảy ra khi \(a=b\).
Vậy \(minS=2\).
\(S=\frac{a^2+b^2}{ab}=\frac{a^2}{ab}+\frac{b^2}{ab}\ge\frac{\left(a+b\right)^2}{2ab}\)( Cauchy-Schwarz dạng Engel )
Lại có : \(2ab\le\frac{\left(a+b\right)^2}{2}\)( AM-GM )
\(\Rightarrow\frac{1}{2ab}\ge\frac{1}{\frac{\left(a+b\right)^2}{2}}=\frac{2}{\left(a+b\right)^2}\Rightarrow\frac{\left(a+b\right)^2}{2ab}\ge2\)
Dấu "=" xảy ra <=> a = b
Vậy MinS = 2
B1
Ta có
\(A=\frac{a^2}{24}+\frac{9}{a}+\frac{9}{a}+\frac{23a^2}{24}\ge3\sqrt[3]{\frac{a^2}{24}.\frac{9}{a}.\frac{9}{a}+\frac{23a^2}{24}}\ge\frac{9}{2}+\frac{23.36}{24}\ge39\)
Dấu "=" xảy ra <=> a=6
Vậy Min A = 39 <=> a=6
\(A=a^2+\frac{18}{a}=a^2+\frac{216}{a}+\frac{216}{a}-\frac{414}{a}\ge3\sqrt[3]{a^2.\frac{216}{a}.\frac{216}{a}}-69=39\)
Đẳng thức xảy ra khi a = 6
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
a) \(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{x^2-1}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{\left(x^2-x\right)+\left(2x^2+2x-3x-3\right)-\left(2x^2-x-3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x}{x+1}\)
MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ
\(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{\left(x^2-1\right)}.\)
\(=\frac{x\left(x-1\right)+\left(2x-3\right)\left(x+1\right)-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
\(\Rightarrow A.B=\frac{x}{\left(x+1\right)}.\frac{x\left(x+1\right)}{\left(x-2\right)}=\frac{x^2}{\left(x-2\right)}=\frac{x^2-4+4}{\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+4}{\left(x-2\right)}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\)
Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :
\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)
\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)
Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)}=\frac{4}{x-2}\Rightarrow x^2+4x+4=4\)
\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)
1)???
2) \(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=2+\dfrac{x^2-4x+4}{x^2-2x+1}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)
Vậy GTNN của A là 2 tại x=2.
3) \(\)Đặt \(a=\dfrac{1}{x+100}\Rightarrow x=\dfrac{1}{a}-100\)
\(D=\dfrac{x}{\left(x+100\right)^2}=a^2x=a^2\left(\dfrac{1}{a}-100\right)=a-100a^2=-100\left(a^2-\dfrac{a}{100}+\dfrac{1}{40000}-\dfrac{1}{40000}\right)=-100\left(a-\dfrac{1}{200}\right)^2+\dfrac{1}{400}\le\dfrac{1}{400}\)
Vậy GTLN của D là \(\dfrac{1}{400}\) tại \(a=\dfrac{1}{200}\Leftrightarrow x=100\)
\(S=a^2+\dfrac{1}{a^2}\)
\(S=\dfrac{1}{16}a^2+\dfrac{1}{a^2}+\dfrac{15}{16}a^2\)
\(S\ge2\sqrt{\dfrac{1}{16}a^2\cdot\dfrac{1}{a^2}}+\dfrac{15}{16}\cdot2^2\)
\(S\ge2\cdot\dfrac{1}{4}+\dfrac{15}{4}\)
\(S\ge\dfrac{17}{4}\)
Vậy \(MINS=\dfrac{17}{4}\Leftrightarrow a=2\)