9900 : 36 - 15 x 11 

= 275 - 165

= 110

9700 : 100 + 36 x 12

= 97 + 432

= 529

( 15792 : 336 ) x 5 + 27 x 11

= 47 x 5 + 297

= 235 + 297

= 532

( 160 x 5 - 25 x 4 ) : 4

= ( 800 - 100 ) : 4

= 700 : 4

= 175

1036 + 64 x 52 - 1827

= 1036 + 3328 - 1827

= 4364 - 1827

= 2537

215 . 869 + 215 . 14 

= 215 . ( 869 + 14 ) 

= 215 . 883 

= 189945 

9900:36-15x11= 110 ;   

9700:100+36x12= 529 ;   

( 15792 : 336 ) x 5 + 27 x 11= 532

(160 x 5 – 25 x 4) : 4= 175

;   1036 + 64 x 52 – 1827 = 2537

   215 x 86 + 215 x 14= 21500

12 054 : (15 + 67) = 147

;   53 x 128 - 43 x 128 = 1280

;   29150 - 136 x 201= 1814

????

?????

A ko có min nha bạn

Giúp mình ik

Ta có: \(4x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(4x^4+3x^2\ge0\forall x\)

=>\(4x^4+3x_{}^2+11\ge11\forall x\)

Dấu '=' xảy ra khi x=0

=>a=11; m=0

\(a^2+m^2=11^2+0^2=121\)

Ta có: \(4x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(4x^4+3x^2\ge0\forall x\)

=>\(4x^4+3x_{}^2+11\ge11\forall x\)

Dấu '=' xảy ra khi x=0

=>a=11; m=0

\(a^2+m^2=11^2+0^2=121\)

\(A\ge11\forall x\)

Dấu '=' xảy ra khi x=0

\(A\ge11\forall x\)

Dấu '=' xảy ra khi x=0

a: Khi x=1 thì \(A=\dfrac{x-8}{x-3}=\dfrac{1-8}{1-3}=\dfrac{-7}{-2}=\dfrac{7}{2}\)

Khi x=2/11 thì \(A=\dfrac{\dfrac{2}{11}-8}{\dfrac{2}{11}-3}=\dfrac{-86}{11}:\dfrac{-31}{11}=\dfrac{86}{31}\)

b: Để A là số nguyên thì \(x-8⋮x-3\)

\(\Leftrightarrow x-3-5⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)