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a) Ta có:A = 6x2 - 6x + 1 = 6(x2 - x + 1/4) - 1/2 = 6(x - 1/2)2 - 1/2
Ta luôn có : (x - 1/2)2 \(\ge\)0 \(\forall\)x --> 6(x - 1/2)2 \(\ge\) 0 \(\)x
=> 6(x - 1/2)2 - 1/2 \(\ge\)-1/2 \(\forall\)x
hay A \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra khi : (x - 1/2)2 = 0 <=> x - 1/2 = 0 <=> x = 1/2
Vậy Amin = -1/2 tại x = 1/2
\(a,A=6x^2-6x+1\)
\(=6\left(x^2-x+\frac{1}{6}\right)\)
\(=6\left[\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+\frac{1}{6}\right]\)
\(=6\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{12}\right]\)
\(=6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(A_{min}=-\frac{1}{12}\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
\(a,A=6x^2-6x+1\)
\(=6\left(x^2-x+\frac{1}{4}\right)-\frac{1}{2}\)
\(=6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
\(b,B=3+2x+3x^2\)
\(=3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{8}{3}\)
\(=3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}\ge\frac{8}{3}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{3}\)
Vậy \(Min_B=\frac{8}{3}\Leftrightarrow x=-\frac{1}{3}\)
\(c,C=4x+2x^2-3\)
\(=2\left(x^2+2x+1\right)-5\)
\(=2\left(x+1\right)^2-5\ge-5\)
Dấu = xảy ra \(\Leftrightarrow x=-1\)
Vậy \(Min_C=-5\Leftrightarrow x=-1\)
\(d,D=10x+6+x^2\)
\(=\left(x^2+10x+25\right)-19\)
\(=\left(x+5\right)^2-19\ge-19\)
Dấu = xảy ra \(\Leftrightarrow x=-5\)
Vậy \(Min_D=-19\Leftrightarrow x=-5\)
\(e,E=8x^2-6x+3\)
\(=8\left(x^2-\frac{3}{4}x+\frac{9}{64}\right)+\frac{15}{8}\)
\(=8\left(x-\frac{3}{8}\right)^2+\frac{15}{8}\ge\frac{15}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{3}{8}\)
Vậy \(Min_E=\frac{15}{8}\Leftrightarrow x=\frac{3}{8}\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
\(D=\left(x^4-2x^3+x^2\right)+\left(2x^2-2x+1\right)\)
\(D=\left(x^2-x\right)^2+2\left(x^2-x\right)+1=\left(x^2-x+1\right)^2\)
\(D=\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]^2\)
\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow D\ge\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)
đẳng thúc khi x=1/2
{logic 10x-->10x^2}
\(E=x^4-6x^3+10x^2-6x+9\)
\(E=\left(x^4-3x+9x^2\right)+\left(x^2-6x+9\right)\)
\(E=\left(x^2-3x\right)^2+\left(x-3\right)^2=\left[x^2\left(x-3\right)^2\right]+\left(x-3\right)^2\)
\(E=\left(x-3\right)^2\left(x^2+1\right)\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(x^2+1\right)\ge1\end{matrix}\right.\) \(\Rightarrow E\ge0\) đẳng thức khi x=3