A =x² -5x +8 >0 với mọi x

= x²-5x+\(\dfrac{25}{4}+\dfrac{7}{4}\)

=\(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)

do \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)

=> \(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

=> A luôn lớn hơn 0 vs mọi x

B= -4x² -4x-2 < 0 với mọi x

=-(4x²+4x+2)

=-4x²-4x-1-1

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

=-\(\left[4\left(x^2+x+\dfrac{1}{4}\right)+1\right]\)

= -\(\left[4\left(x+\dfrac{1}{2}\right)^2+1\right]\)

=-4\(\left(x+\dfrac{1}{2}\right)^2-1\)

do \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)

=> -4 \(\left(x+\dfrac{1}{2}\right)^2\le0\)

=> \(-4\left(x+\dfrac{1}{2}\right)^2-1\le-1\)

vậy B luôn nhỏ hơn 0 vs mọi x

a) ($x$ + 19) $⋮$ ($x$ + 1)

=> \(\left(x+19\right)-\left(x+1\right)⋮\left(x+1\right)\)

=> \(\left(x+19-x-1\right)⋮\left(x+1\right)\)

=> \(18⋮\left(x-1\right)\)

=> \(\left(n+1\right)\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

ta có bảng sau

vậy x\(\in\left\{0;1;2;5;8;17\right\}\)

\(4n⋮\left(n-1\right)\)

vì \(\left(n-1\right)⋮\left(n-1\right)\)

=>\(4\left(n-1\right)⋮\left(n-1\right)\)

=> \(\left(4n-4\right)⋮\left(n-1\right)\)

=> \(4n-\left(4n-4\right)⋮\left(n-1\right)\)

=> \(\left(4n-4n+4\right)⋮\left(n-1\right)\)

=> \(4⋮\left(n-1\right)\)

=> \(n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

ta có bảng sau

vậy n\(\in\left\{-3;-1;0;2;3;5\right\}\)

\(2n+3⋮n+4\)

vì \(\left(n+4\right)⋮\left(n+4\right)\)

=>\(2\left(n+4\right)⋮\left(n+4\right)\)

=>\(\left(2n+8\right)⋮\left(n+4\right)\)

=> \(\left(2n+3\right)-\left(2n+8\right)⋮\left(n+4\right)\)

=> \(\left(2n+3-2n-8\right)⋮\left(n+4\right)\)

=> \(-5⋮\left(n+4\right)\)

\(\left(n+4\right)\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)

ta có bảng sau

vậy x\(\in\left\{-9;-5;-3;1\right\}\)

B1:lấy nước đầy can 4 lít rồi đổ vào can 5 lít

B2:lấy nước đầy can 4 lít lại đổ vào can 5 lít(lúc này can 5 lít đã có 4 lít nước)cho đầy số nước còn lại ở trong bình 4 lít có 3 lít

để a97b \(⋮5\) thì b=0 hoặc b=5

để \(a970⋮9\) thì a+9+7+0 \(⋮9\) => a=2

Để \(a975⋮9\) thì a+9+7+5\(⋮9\) => a=6

=> a=2 thì b=0

a=6 thì b=6

2n+ 18 \(⋮\) 2n+5

=> \(\left(2n+18\right)-\left(2n+5\right)⋮\left(2n+5\right)\)

=> \(\left(2n+18-2n-5\right)⋮\left(2n+5\right)\)

=> \(13⋮\left(2n+5\right)\)

=> \(\left(2n+5\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

ta có bảng sau

vây n \(\in\left\{-9;-3;-2;4\right\}\)

A=5+6x-x²

= x²+6x+5

= x²+6x+9-4

= (x²+6x+9)-4

= (x+3)²-4

do (x+3)²\(\ge0\forall x\)

=> (x+3)²-4\(\ge-4\)

= > A \(\ge-4\)

GTLN A =-4 khi

x+3=0

=>x=-3

vậy GTLN a=-4khi x=-3

A= 3x² - 12x+10

= 3x²-12x+12-2

= 3(x²-4x+4)-2

=3(x-2)²-2

do 3(x-2)²\(\ge0\forall x\)

=> 3(x-2)²-2\(\ge-2\)

=> A\(\ge-2\)

Min A =-2 khi

x-2=0

=> x=2

A= 3x² - 12x+10

=3x²-12x+12-2

=3(x²-4x+4)-2

=3(x-2)²-2

do3(x-2)2\(\ge0\forall x\)

=> 3(x-2)2-2\(\ge-2\)

=> A\(\ge-2\)

Min A=-2khi

x-2=0

=> x=2