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MS
1
2 tháng 8 2021
\(D=3x^2+2x+1\)
\(D=\left(3x^2+2x+\frac{\sqrt{3}}{3}^2\right)+\frac{2}{3}\)
\(D=\left(\sqrt{3}x+\frac{\sqrt{3}}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
dấu "=" xảy ra khi và chỉ khi
\(x=\frac{1}{3}\)
\(< =>MIN:D=\frac{2}{3}\)
MS
2
XO
2 tháng 8 2021
Ta có : \(D=3x^2+2x+1=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)=3\left(x^2+\frac{2}{3}x+\frac{1}{9}+\frac{2}{9}\right)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
\(\Rightarrow\)Min D = 2/3
Dấu "=" xảy ra khi x + 1/3 = 0
\(\Rightarrow x=-\frac{1}{3}\)
Vậy Min D = 2/3 khi x = -1/3
2 tháng 8 2021
D = 3x2 + 2x + 1 = 3( x2 + 2/3x + 1/9 ) + 2/3 = 3( x + 1/3 )2 + 2/3 ≥ 2/3 ∀ x
Dấu "=" xảy ra <=> x = -1/3 . Vậy MinD = 2/3
22 tháng 1
a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)
=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)
=>18x-12>=12x+12
=>6x>=24
=>x>=4
b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)
=>\(x^2+2x+1< x^2-2x+1\)
=>4x<0
=>x<0
c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì
\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)
=>\(2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
=>x<=4
HH
2
27 tháng 1 2022
3) Ta có: \(C=x^2-4x+7=\left(x-2\right)^2+3\ge3\)
Dấu "=" xảy ra khi x = 2
4) Ta có: \(D=2x^2+3x+4=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{23}{8}=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{23}{8}\ge\dfrac{23}{8}\)
Dấu "=" xảy ra khi \(x=-\dfrac{3}{4}\)
MH
27 tháng 1 2022
3) \(C=x^2-4x+7\)
\(=\left(x-2\right)^2+3\text{≥}3\) ∀x (vì \(\left(x-2\right)^2\text{≥}0\))
MinC=3 ⇔ x=2
4) \(D=2x^2+3x+4\)
\(=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{23}{8}\text{≥}\dfrac{23}{8}\) ∀x (vì \(2\left(x+\dfrac{3}{4}\right)^2\text{≥}0\))
MinD= \(\dfrac{23}{8}\) ⇔ \(x=-\dfrac{3}{4}\)
YN
21 tháng 12 2021
Answer:
a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)
\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)
\(\Rightarrow5x+2x+2-12=0\)
\(\Rightarrow7x-10=0\)
\(\Rightarrow x=\frac{10}{7}\)
b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)
\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)
\(\Rightarrow\frac{3}{2}x=-6\)
\(\Rightarrow x=-4\)
c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)
\(\Rightarrow9x-6-6x-6\ge0\)
\(\Rightarrow3x-12\ge0\)
\(\Rightarrow x\ge4\)
d) \(\left(x+1\right)^2< \left(x-1\right)^2\)
\(\Rightarrow x^2+2x+1< x^2-2x+1\)
\(\Rightarrow4x< 0\)
\(\Rightarrow x< 0\)
e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)
\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)
\(\Rightarrow6x\le24\)
\(\Rightarrow x\le4\)
f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)
\(\Rightarrow9x-6-6x-6\le0\)
\(\Rightarrow3x\le12\)
\(\Rightarrow x\le4\)
H
1
10 tháng 8 2021
a) Ta có: \(A=x^2-3x+5\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
b: Ta có: \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5\ge5\forall x\)
Dấu '=' xảy ra khi x=0
`S=3x^2+2x+1`
`=(3x^2+2x+1/3)+2/3`
`=[(\sqrt3 x)^2+ 2.\sqrt3 x . 1/\sqrt3 + (1/\sqrt3)^2]+2/3`
`=(\sqrt3 x+1/\sqrt3)^2 + 2/3`
`=(\sqrt3x+\sqrt3/3)^2+2/3`
`=> D_(min) =2/3 <=> \sqrt3x+\sqrt3/3=0 <=>x=-1/3`
\(D=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)=3\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{2}{9}\right)=3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)
dấu"=" xảy ra<=>x=-1/3