Bài 2: 

a: =>x-65=-20

hay x=45

1/4

11

1/4

11

a: f(-2)-g(1/2)

\(=5\left(-2\right)-3+4\cdot\dfrac{1}{2}-1\)

\(=-10-4+2=-10-2=-12\)

b: \(2\cdot f^2\left(-3\right)-3\cdot g^2\left(-2\right)\)

\(=2\cdot\left[5\cdot\left(-3\right)-3\right]^2-3\cdot\left[\left(-4\right)\left(-2\right)+1\right]^2\)

\(=2\cdot\left(-18\right)^2-3\cdot9^2\)

\(=648-3\cdot81=405\)

A = 78 x 31 + 78 x 24 + 78 x 17 + 22 x 72

A = 78 x ( 31 + 24 + 17 ) + 22 x 72

A = 78 x 72 + 22 x 72

A = 72 x ( 78 + 22 ) 

A = 72 x 100

A = 7200

C = 53 x 39 + 47 x 39 - 53 x 21 - 47 x 21

C = 39 x ( 53 + 47 ) - 21 x ( 53 - 47 )

C = 39 x 100 - 21 x 6

C = 3900 - 126

C = 3774

ko pc tui thua bn 1 lp 

A = 78 x 31 + 78 x 24 + 78 x 17 + 22 x 72

A = 78 x ( 31 + 24 +17 ) + 22 x 72

A = 78 x 72 + 22 x 72

A = 72 x ( 78 + 22 )

A = 72 x 100

A = 7200

4:

a: f(x)=0

=>-x-4=0

=>x=-4

b: g(x)=0

=>x^2+x+4=0

Δ=1^2-4*1*4=1-16=-15<0

=>g(x) ko có nghiệm 

c: m(x)=0

=>2x-2=0

=>x=1

d: n(x)=0

=>7x+2=0

=>x=-2/7

a) =7
b) =16
c) =69

189 : (3 x 9)                    672 : (6 x 7)

= 189 : 27                      = 672 : 42

= 7                                 = 16            

1656 : (3 x 8) 

= 1656 : 24

= 69

a. 189 : (3 x 9) 

= 189 : 27 = 7.

b. 672 : (6 x 7)

= 672 : 42 = 16.

c. 1656 : (3 x 8)

= 1656 : 24 = 69.

a: \(=15x^3-6x^2-3x\)

e: \(=x^2-12x+35\)

a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)

\(\Leftrightarrow5⋮\left(x+1\right)\)

mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)

Vậy...

b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)

\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)

Vậy...

c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)

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

Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

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

Vậy...

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

Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)

Vậy...