(-25) . 1 + 72 . (-25) + (-25) . 27

=(-25) . ( 1+72+27)

=(-25) . 100

=  -2500

( - 25 ) + 72 . ( - 25 ) + ( - 25 ) . 27 

= ( - 25 ) . ( 1 + 72 + 27 ) 

= ( - 25 ) . 100 

= - 2500 

Đáp án là -2500

Mình không biết giải từng bước thôi

bài 2 tìm x

a,106- ( x+ 7) =9

x+7 = 106 - 9

x+7 = 107

x= 107 - 7

x=100

b, 2 x ( x+ 4) + 5 =65

2 x (x+4) = 65 - 5

2 x (x+4) = 60

x+4 = 60:2

x+4= 30

x= 30 - 4

x=26

c, (16x x -32) x 45=0

16 x X - 32 = 0: 45

16 x X - 32 =0

16 x X = 0 + 32

16 x X = 32

X= 32:16

X=2

d, x+4 x x = 100 : 5

X + 4 x X = 20

(1+4) x X = 20

5 x X  = 20

X= 20:5

X=4

Bài 1:

a, 24 x 66 - 24 + 36 x 24

= 24 x (66-1+36)

= 24 x 101

= 24 x 100 + 24

= 2400+24

=2424

b, 29 x 173 - 29 x 72 - 29

= 29 x (173 - 72 -1)

= 29 x 100

= 2900

c, 1+5+10+15+...+55+60

= (5+60) x 6 + 1 (có 6 cặp)

= 65 x 6 + 1

= 390 +1 =391

d, D= 1 x 2 + 2 x 3 + 3 x 4 +...+18 x 19 + 19 x 20

3D= 1 x 2 x 3+ 2 x 3 x (4-1) + 3 x 4 x (5-2) +...+ 18 x 19 x (20 - 17) + 19 x 20 x (21-18)

3D= 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 +.... + 19 x 20 x 21 - 18 x 19 x 20

3D= 19 x 20 x 21 =7980

=> D= 7980:3=2660

\(a,=\dfrac{12x^2y\left(x+y\right)^2:3x\left(x+y\right)}{3x\left(x+y\right):3x\left(x+y\right)}=4xy\left(x+y\right)\\ b,=\dfrac{x^2+6x+9+24-x^2+4x-3}{4\left(x-3\right)\left(x+3\right)}=\dfrac{10\left(x+3\right)}{4\left(x-3\right)\left(x+3\right)}=\dfrac{4}{2\left(x-3\right)}\)

Ko có j là ko thể vs...máy tính!

a/ 90201/7

b/ 2164

bai toan nay kho

Tính lần lượt ra

a)

ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)

Ta có: \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)

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

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

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

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

\(=\dfrac{-x-3}{x}\)

b) Để A nguyên thì \(-x-3⋮x\)

mà \(-x⋮x\)

nên \(-3⋮x\)

\(\Leftrightarrow x\inƯ\left(-3\right)\)

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

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-1\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{1;-1\right\}\)

a: 3/4=18/24

9/24=9/24

b: 7/8=35/40

12/40=12/40

c: 9/27=9/27

3/9=9/27

a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)

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

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

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

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

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

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

\(=\left(x-2\right)\left(x+2+x-2\right)\)

\(=\left(x-2\right)2x\)

c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

a) => x².(x-3)-4(x-3)=(x-3)(x²-4)=(x-3)(x-2)(x+2)

b) => (x+2)(x-2)+(x-2)²=(x-2)(x+2+x-2)=2x(x-2)

c) => x³+27-(4x²+12x)=(x+3)(x²-3x+3)-4x(x+3)=(x+3)(x²-3x+3-4x)=(x-3)(x²-7x+3)