e,E= 25x^2 – 10x – 4 f,F= x^2 + x + 1 g, G=x^2 + 3x – 5 h,H= x^2 – 5x + 3 22/08/2021 Bởi Madelyn e,E= 25x^2 – 10x – 4 f,F= x^2 + x + 1 g, G=x^2 + 3x – 5 h,H= x^2 – 5x + 3
e) E= $(5x)^{2}$ -10x+1 -5 = $(5x-1)^{2}$ -5 Vì $(5x-1)^{2}$ ≥0 ∀x ⇒$(5x-1)^{2}$ -5 ≥-5 ∀x E = -5 ⇔ 5x-1=0 ⇔x=$\frac{1}{5}$ Vậy MinE=-5 ⇔ x=$\frac{1}{5}$ f) F= $x^{2}$ +x+1 = $x^{2}$ + 2$\frac{1}{2}$x +$\frac{1}{4}$+ $\frac{3}{4}$ = $(x+\frac{1}{2})^{2}$ +$\frac{3}{4}$ Vì $(x+\frac{1}{2})^{2}$ ≥0 ∀x ⇒ $(x+\frac{1}{2})^{2}$ +$\frac{3}{4}$ ≥ $\frac{3}{4}$ ∀x hay F ≥ $\frac{3}{4}$ F = $\frac{3}{4}$ ⇔ x+$\frac{1}{2}$ =0 ⇔ x= – $\frac{1}{2}$ Vậy MinF=$\frac{3}{4}$ ⇔ x= – $\frac{1}{2}$ g) G= $x^{2}$ +2.$\frac{3}{2}$x+ $\frac{9}{4}$ – $\frac{29}{4}$ = $(x+\frac{3}{2})^{2}$ – $\frac{29}{4}$ Vì $(x+\frac{3}{2})^{2}$ ≥0 ∀x ⇒ $(x+\frac{3}{2})^{2}$ – $\frac{29}{4}$ ≥- $\frac{29}{4}$ ∀x hay G ≥- $\frac{29}{4}$ G =- $\frac{29}{4}$ ⇔ x+$\frac{3}{2}$ =0 ⇔ x = -$\frac{3}{2}$ Vậy MinG = – $\frac{29}{4}$ ⇔ x = -$\frac{3}{2}$ h) H= $x^{2}$ -2.$\frac{5}{2}$x+ $\frac{25}{4}$ -$\frac{13}{4}$ =$(x-\frac{5}{2})^{2}$ -$\frac{13}{4}$ Vì $(x-\frac{5}{2})^{2}$ ≥0 ∀x ⇒$(x-\frac{5}{2})^{2}$ -$\frac{13}{4}$ ≥ -$\frac{13}{4}$ ∀x hay H ≥ $\frac{13}{4}$ H = – $\frac{13}{4}$ ⇔ x-$\frac{5}{2}$ =0 ⇔ x= $\frac{5}{2}$ Vậy MinH = -$\frac{13}{4}$ ⇔x= $\frac{5}{2}$ Bình luận
Giải thích các bước giải: e) $E= 25x^2-10x-4$ $=(25x^2-10x+1)-5\\=(5x-1)^2-5$ Có `(5x-1)^2>=0` `=>(5x-1)^2-5>=-5` `=>E>=-5` Dấu “=” xảy ra `<=>5x-1=0=>5x=1=>x=1/5` Vậy `Emin=-5<=>x=1/5.` f) `F= x^2+x+1` `=(x^2+x+1/4)+3/4` `=(x+1/2)^2+3/4` Có `(x+1/2)^2>=0` `=>(x+1/2)^2+3/4>=3/4` `=>F>=3/4` Dấu “=” xảy ra `<=>x+1/2=0=>x=-1/2` Vậy `Fmin=3/4<=>x=-1/2.` g) `G=x^2+3x-5` `=(x^2+3x+9/4)-29/4` `=(x+3/2)^2-29/4` Có `(x+3/2)^2>=0` `=>(x+3/2)^2-29/4>=-29/4` `=>G>=1` Dấu “=” xảy ra `<=>x+3/2=0=>x=-3/2` Vậy `Gmin=-29/4<=>x=-3/2.` h) `H=x^2-5x+3` `=(x^2-5x+25/4)-13/4` `=(x-5/2)^2-13/4` Có `(x-5/2)^2>=0` `=>(x-5/2)^2-13/4>=-13/4` `=>H>=-13/4` Dấu “=” xảy ra `<=>x-5/2=0=>x=5/2` Vậy `Hmin=-13/4<=>x=5/2.` Bình luận
e) E= $(5x)^{2}$ -10x+1 -5
= $(5x-1)^{2}$ -5
Vì $(5x-1)^{2}$ ≥0 ∀x
⇒$(5x-1)^{2}$ -5 ≥-5 ∀x
E = -5 ⇔ 5x-1=0
⇔x=$\frac{1}{5}$
Vậy MinE=-5 ⇔ x=$\frac{1}{5}$
f) F= $x^{2}$ +x+1
= $x^{2}$ + 2$\frac{1}{2}$x +$\frac{1}{4}$+ $\frac{3}{4}$
= $(x+\frac{1}{2})^{2}$ +$\frac{3}{4}$
Vì $(x+\frac{1}{2})^{2}$ ≥0 ∀x
⇒ $(x+\frac{1}{2})^{2}$ +$\frac{3}{4}$ ≥ $\frac{3}{4}$ ∀x
hay F ≥ $\frac{3}{4}$
F = $\frac{3}{4}$ ⇔ x+$\frac{1}{2}$ =0
⇔ x= – $\frac{1}{2}$
Vậy MinF=$\frac{3}{4}$ ⇔ x= – $\frac{1}{2}$
g) G= $x^{2}$ +2.$\frac{3}{2}$x+ $\frac{9}{4}$ – $\frac{29}{4}$
= $(x+\frac{3}{2})^{2}$ – $\frac{29}{4}$
Vì $(x+\frac{3}{2})^{2}$ ≥0 ∀x
⇒ $(x+\frac{3}{2})^{2}$ – $\frac{29}{4}$ ≥- $\frac{29}{4}$ ∀x
hay G ≥- $\frac{29}{4}$
G =- $\frac{29}{4}$ ⇔ x+$\frac{3}{2}$ =0
⇔ x = -$\frac{3}{2}$
Vậy MinG = – $\frac{29}{4}$ ⇔ x = -$\frac{3}{2}$
h) H= $x^{2}$ -2.$\frac{5}{2}$x+ $\frac{25}{4}$ -$\frac{13}{4}$
=$(x-\frac{5}{2})^{2}$ -$\frac{13}{4}$
Vì $(x-\frac{5}{2})^{2}$ ≥0 ∀x
⇒$(x-\frac{5}{2})^{2}$ -$\frac{13}{4}$ ≥ -$\frac{13}{4}$ ∀x
hay H ≥ $\frac{13}{4}$
H = – $\frac{13}{4}$ ⇔ x-$\frac{5}{2}$ =0
⇔ x= $\frac{5}{2}$
Vậy MinH = -$\frac{13}{4}$ ⇔x= $\frac{5}{2}$
Giải thích các bước giải:
e) $E= 25x^2-10x-4$
$=(25x^2-10x+1)-5\\=(5x-1)^2-5$
Có `(5x-1)^2>=0`
`=>(5x-1)^2-5>=-5`
`=>E>=-5`
Dấu “=” xảy ra `<=>5x-1=0=>5x=1=>x=1/5`
Vậy `Emin=-5<=>x=1/5.`
f) `F= x^2+x+1`
`=(x^2+x+1/4)+3/4`
`=(x+1/2)^2+3/4`
Có `(x+1/2)^2>=0`
`=>(x+1/2)^2+3/4>=3/4`
`=>F>=3/4`
Dấu “=” xảy ra `<=>x+1/2=0=>x=-1/2`
Vậy `Fmin=3/4<=>x=-1/2.`
g) `G=x^2+3x-5`
`=(x^2+3x+9/4)-29/4`
`=(x+3/2)^2-29/4`
Có `(x+3/2)^2>=0`
`=>(x+3/2)^2-29/4>=-29/4`
`=>G>=1`
Dấu “=” xảy ra `<=>x+3/2=0=>x=-3/2`
Vậy `Gmin=-29/4<=>x=-3/2.`
h) `H=x^2-5x+3`
`=(x^2-5x+25/4)-13/4`
`=(x-5/2)^2-13/4`
Có `(x-5/2)^2>=0`
`=>(x-5/2)^2-13/4>=-13/4`
`=>H>=-13/4`
Dấu “=” xảy ra `<=>x-5/2=0=>x=5/2`
Vậy `Hmin=-13/4<=>x=5/2.`