a, \(A=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)

=> \(A\ge-\frac{9}{4}\) dấu = xảy ra khi : \(x=\frac{-1}{2}\)

b, \(B=x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

=> \(B\ge-\frac{1}{4}\) dấu = <=> \(x=\frac{1}{2}\)

Bn gửi từng câu sẽ có nhều ng trl hơn nhé

tý mk giải câu a cho cần ko

Ta có : A = 9x² - 6x + 2 

= 9x² - 6x + 1 + 1 = (3x - 1)² + 1 \(\ge\)1 

=> Min A = 1

Dấu "=" xảy ra <=> 3x - 1 = 0 

<=> x = 1/3

Vậy Min A = 1 <=> x = 1/3

b) Ta có 2B = 4x² + 4x + 2 

= 4x² + 4x + 1 + 1 

= (2x + 1)² + 1 \(\ge\)1

=> B \(\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> 2x + 1 = 0 

<=> x = -1/2

Vậy Min B = 1/2 <=> x = -1/2

c) C = (2x - 1)² + (x - 2)² 

= 5x² - 8x + 5

=> 5C = 25x² - 40x + 25 

 = 25x² - 40x + 16 + 9 

= (5x - 4)² + 9 \(\ge9\)

=> \(C\ge\frac{9}{5}\)

Dấu "=" xảy ra <=> 5x - 4 = 0 

<=> x = 0,8

Vậy Min C = 9/5 <=> x = 0,8

d) D = 3x² + 5x = \(3\left(x^2+\frac{5}{3}x\right)=3\left(x^2+2.\frac{5}{6}x+\frac{25}{36}-\frac{25}{36}\right)=3\left(x+\frac{5}{6}\right)^2-\frac{25}{12}\ge-\frac{25}{12}\)

=> \(D\ge-\frac{25}{12}\)

Dấu "=" xảy ra <=> x + 5/6 = 0 

<=> x = -5/6

Vậy Min D = -25/12 <=> x = -5/6e) E = (x -2)(x - 3)(x + 5)x

= (x² - 5x + 6)(x² + 5x)

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau

a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

=>1+3x-6=3-x

=>3x-5=3-x

=>4x=8

hay x=2(loại)

b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)

=>8-x-8x+56=-26

=>-9x+64=-26

=>-9x=-90

hay x=10(nhận)

c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)

\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)

=>-7x+10x=12-5

=>3x=7

hay x=7/3(nhận)