1. a, => -12x+60+21-7x = 5

=> 81 - 19x = 5

=> 19x = 81 - 5 = 76

=> x = 76 : 19 = 4

Tk mk nha

a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)

b/

1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Suy ra Min A = 7 <=> x = 2

2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Suy ra Min B = 1/4 <=> x = 1/2

3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)

\(\ge-\frac{9}{2}\)

Suy ra Min N = -9/2 <=> x = 1/2

\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)

1.A=|7-x|-2

Vì |7-x| luôn luôn > hoặc bằng 0

⇒ Để A min =-2 khi và chỉ khi 7-x=0⇒x=7

Vậy để Amin =-2 ⇔x=7

2 B=-|5+x|+10

Vì -|5+x| luôn luôn , hoặc bằng 0

⇒ Để Bmax =10 khi và chỉ khi 5+x=0⇒x=-5

3 D =(x-4)^2-1

Vì (x-4)^2 luôn luôn > hoặc bằng0

Để Dmin =-1 khi và chỉ khi x-4=0⇒x=4

`|7-x|>=0`

`=>A>=-2`

Dấu "=" xảy ra khi `x=7`

`-|5+x|<=0`

`=>B<=10`

Dấu "=" xảy ra khi `x=-5`

`(x-4)^2>=0`

`=>C>=-1`

Dấu "=" xảy ra khi `x=4`