a/ (x-1)²-(4x+3)(2-x)=x²-2x+1-(8x-4x²+6-3x)

=x²-2x+1-8x+4x²-6+3x=5x²-7x-6

b/ (15x³y² - 6x²y³) : 3x²y² = 5x - 2y

c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)

bài2

A= -x² + 6x + 12

=-(x²-6x-12)

=-(x²-6x+9+3)

= -(x-3)²+3

do -(x-3)2\(\le0\forall x\)

=> -(x-3)2+3\(\le3\)

GTLN A =3 khi và chỉ khi

x-3=0

=> x=3

vậy GTLN A =3khi x=3

\(a,A=x^2+15x-25\)

\(=x^2+2.x.\frac{15}{2}+\frac{125}{4}-\frac{125}{4}-25\)

\(=\left(x+\frac{15}{2}\right)^2-\frac{225}{4}\)

\(A_{min}=-\frac{225}{4}\Leftrightarrow\left(x+\frac{15}{2}\right)^2=0\)

\(\Leftrightarrow x=-\frac{15}{2}\)

1

a, 4x - 3x + 1 = 5

        x             =5-1

       x              =4

Vậy x=4

b,  (2x - 4 ) . 3x =0

=> 2x - 4 =0     hoặc  3x = 0

=> 2x       =4    hoặc    x=0

=>  x        =2    hoặc    x=0

vậy x= 2 hoặc x=0

c, x . ( x -1 ) - ( x-1 )=0

    (x-1) . (x-1 )         =0

     (x-1)²                  =0²

     x-1                      =0 

    x                          =1

vậy x=1 

2/ a,  7 . (x - 1 )  = 6x + 3

         7x  -7        = 6x  +3

         7x - 6x       =7+3 

              x            =10

vậy x=10 

b, 8 . ( 2x - 3 )  -15x   =4

    16x  - 24       -15x   =4

     16x - 15x                =4+24

            x                      =28

vậy x=28

c, 7 . 10 + ( x-1 ) .2    =100

      70    +   2x  -2      =100

                    2x -2      =100-70

                    2x -2      =30 

                    2x           =30+2

                    2x           =32

                      x           =16

vậy x=16

chúc bn học tốt

a, \(A=9x^2-6x+5\)

\(=\left(9x^2-6x+1\right)+4\)

\(=\left(3x-1\right)^2+4\)

ta có:

\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)

Vậy Min A = 4

Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(b,B=4x^2-5x\)

\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)

\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)

TA có:

\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)

Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)

\(c,C=3x^2-6x\)

\(=3\left(x^2-2x+1\right)-3\)

\(=3\left(x-1\right)^2-3\)

Ta có:

\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)

vậy Min C = -3

Để C = -3 thì x-1=0 => x = 1

\(d,D=5x^2-15x\)

\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)

\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)

Ta có:

\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)

Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(e,E=x^2+3x+4\)

\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(f,F=2x^2-4x+7\)

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

\(=2\left(x-1\right)^2+5\ge5\forall x\)

Vậy Min F = 5 khi x - 1 =0 => x = 1

\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)

\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)

Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)

\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)

\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)

Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

\(A=3x^2+6x+7\)

\(A=\left(3x^2+6x+3\right)+4\)

\(A=3\left(x^2+2x+1\right)+4\)

\(A=3\left(x+1\right)^2+4\)

\(\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+4\ge4\)

\(\Rightarrow A\ge4\)

dấu "=" xảy ra khi : 

(x + 1)² = 0 => x + 1 = 0 => x = -1

vậy gtnn của A = 4 khi x = -1

bạn tự kết luận nhé

\(A=x^2-6x+13=\left(x-3\right)^2+4\ge4\)

\(B=3x^2-6x+3=3\left(x-1\right)^2\ge0\)

\(C=x^2-2x+1+1=\left(x-1\right)^2+1\ge1\)

\(D=x^2+6x+9=\left(x+3\right)^2\ge0\)

      B = (x-2)(x-5)(x²-7x-10)

    =(x²-7x+10)(x²-7x-10)

    =(x²-7x)²-10²

     ⁼(x²-7x)²-100

=>GTNN của B là 100 <=>x²-7x=0

             x(x-7)=0

        =>x=0 hoặc x=7

Vậy GTNN của B là 100 khi x=0 hoặc x=7

A=x^2+2x.3/2+3/2^2+11/2

=(x+3/2)^2+11/2>=11/2

hình như linh chi làm sai r

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)