câu 1 :tìm gtnn
A=x mủ 2 -4xy+5y mủ 2+10x-22y+28
B=2x mủ 2 -6x
câu 2 tìm gtln
A=4x-2x mủ 2 +3
B=x-x mủ 2
C=2x-2x mủ 2 -5
lưu ý : áp dụng hằng đẳng thức đáng nhớ
câu 1 :tìm gtnn A=x mủ 2 -4xy+5y mủ 2+10x-22y+28 B=2x mủ 2 -6x câu 2 tìm gtln A=4x-2x mủ 2 +3 B=x-x mủ 2 C=2x-2x mủ 2 -5 lưu ý : áp dụng hằng đẳng t
By Kaylee
Đáp án:
`1)A=x^2-4xy+5y^2+10x-22y+28`
`=x^2-4xy+4y^2+10(x-2y)+y^2-2y+28`
`=(x-2y)^2+10(x-2y)^2+25+y^2-2y+1+2`
`=(x-2y+5)^2+(y-1)^2+2>=2`
Dấu “=” xảy ra khi \(\begin{cases}y=1\\x=2y-5=-3\\\end{cases}\)
`B=2x^2-6x`
`=2(x^2-3x)`
`=2(x^2-2*x*3/2+9/4-9/4)`
`=2(x-3/2)^2-9/2>=-9/2`
Dấu “=” xảy ra khi `x=3/2`
`2)A=4x-2x^2+3`
`=-2(x^2-2x)+3`
`=-2(x^2-2x+1-1)+3`
`=-2(x-1)^2+5<=5`
Dấu “=” xảy ra khi `x=1`
`B=x-x^2`
`=-(x^2-x)`
`=-(x^2-2*x*1/2+1/4-1/4)`
`=-(x-1/2)^2+1/2<=1/2`
Dấu “=” xảy ra khi `x=1/2`
`C=2x-2x^2-5`
`=-2(x^2-x)-5`
`=-2(x^2-2*x*1/2+1/4-1/4)-5`
`=-2(x-1/2)^2+1/2-5`
`=-2(x-1/2)^2-9/2<=-9/2`
Dấu “=” xảy ra khi `x=1/2.`
Câu `1:`
`a) A = x^2 – 4xy + 5y^2 + 10x – 22y +28`
` = x^2 – 4xy + 4y^2 + y^2 + 10x – 20y – 2y + 25 + 2 + 1`
` = (x^2 – 4xy + 4y^2) + (10x – 20y) + 25 + (y^2 – 2y + 1) + 2`
` = [ (x-2y)^2 + 2 . 5 . (x-2y) + 5^2 ] + (y-1)^2 + 2`
` = (x- 2y + 5)^2 + (y-1)^2 + 2`
`\forall x ; y` ta có :
`(x-2y+5)^2 \ge 0`
`(y-1)^2 \ge 0`
`=> (x- 2y + 5)^2 + (y-1)^2 \ge 0`
`=> (x- 2y + 5)^2 + (y-1)^2 + 2 \ge 2`
`=> A \ge 2`
Dấu `”=”` xảy ra `<=>`$\begin{cases} y – 1 = 0 \\ x – 2y + 5 = 0 \end{cases}$
`<=>` $\begin{cases} y = 1 \\ x = -3 \end{cases}$
Vậy `\text{Min}_A = 2 <=>` $\begin{cases} y = 1 \\ x = -3 \end{cases}$
`b) B = 2x^2 – 6x`
` = 2 . (x^2 – 3x)`
` = 2 . (x^2 – 3x + 9/4) – 9/2`
` = 2 . (x – 3/2)^2 – 9/2`
`\forall x` ta có :
`(x-3/2)^2 \ge 0`
`=> 2 . (x-3/2)^2 \ge 0`
`=> 2 . (x-3/2)^2 – 9/2 \ge -9/2`
`=> B \ge -9/2`
Dấu `”=”` xảy ra `<=> x – 3/2 = 0 <=> x = 3/2`
Vậy `\text{Min}_B = -9/2 <=> x = 3/2`
Bài `2:`
`a) A = 4x – 2x^2 + 3`
` = – (2x^2 – 4x) + 3`
` = -2. (x^2 + 2x) + 3`
` = -2 . (x^2 + 2x +1) +5`
` = -2 . (x+1)^2 + 5`
`\forall x` ta có :
`(x+1)^2 \ge 0`
`=> -2 . (x+1)^2 \le 0`
`=> -2 . (x+1)^2 + 5 \le 5`
`=> A \le 5`
Dấu `”=”` xảy ra `<=> x + 1 = 0 <=> x = -1`
Vậy `\text{Max}_A = 5 <=> x = -1`
`b) B = x – x^2`
` = – (x^2 – x)`
` = – (x^2 – x + 1/4) + 1/4`
` = -(x-1/2)^2 + 1/4`
`\forall x` ta có :
`(x-1/2)^2 \ge 0`
`=> -(x – 1/2)^2 \le 0`
`=> -(x-1/2)^2 + 1/4 \le 1/4`
`=> B \le 1/4`
Dấu `”=”` xảy ra `<=> x-1/2 =0<=> x=1/2`
Vậy `\text{Max}_B = 1/4 <=> x = 1/2`
`c) C = 2x – 2x^2 – 5`
` = -(2x^2 – 2x) – 5`
` = -2.(x^2 – x) – 5`
` = -2 . (x^2 – x + 1/4) – 9/2`
` = -2. (x-1/2)^2 – 9/2`
`\forall x` ta có :
`(x-1/2)^2 \ge 0`
`=> -2 . (x-1/2)^2 \le 0`
`=>-2 . (x-1/2)^2 – 9/2 \le -9/2`
`=> C \le -9/2`
Dấu `”=”` xảy ra `<=> x – 1/2 = 0 <=> x = 1/2`
Vậy `\text{Max}_C = -9/2 <=> x = 1/2`