Ta có:

\(56=2^3\cdot7\\ 84=2^2\cdot3\cdot7\\ 120=2^3\cdot5\cdot3\\=>ƯCLN\left(56;84;120\right)=2^2=4\)

\(2^{x+2}+2^{x+2}+2^{x+1}=224\\ =>2^{x+1}\cdot2+2^{x+1}\cdot2+2^{x+1}\cdot1=224\\ =>2^{x+1}\cdot\left(2+2+1\right)=224\\ =>2^{x+1}\cdot5=224\\ =>2^{x+1}=\dfrac{224}{5}\\ =>x+1=log_2\dfrac{224}{5}\\ =>x=log_2\dfrac{224}{5}-1\)

\(1,x^2+9-16y^2+6x\\ =\left(x^2+6x+9\right)-\left(4y\right)^2\\ =\left(x+3\right)^2-\left(4y\right)^2\\ =\left(x-4y+3\right)\left(x+4y+3\right)\\ 2,x^2-9+y^2+2xy\\ =\left(x^2+2xy+y^2\right)-9\\ =\left(x+y\right)^2-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ 3,x^2-4x+4-9y^2\\ =\left(x-2\right)^2-\left(3y\right)^2\\ =\left(x-3y-2\right)\left(x+3y-2\right)\\ 4,x^2-4xy+4y^2-81\\ =\left(x-2y\right)^2-9^2\\ =\left(x-2y-9\right)\left(x-2y+9\right)\\ 5,6x^2+6y^2-24+12xy\\ =\left(6x^2+12xy+6y^2\right)-24\\ =6\left[\left(x^2+2xy+y^2\right)-4\right]\\ =6\left[\left(x+y\right)^2-2^2\right]\\ =6\left(x+y-2\right)\left(x+y+2\right)\\ 6,9x^2-6x+1-25\\ =\left(3x-1\right)^2-5^2\\ =\left(3x-1-5\right)\left(3x-1+5\right)\\ =\left(3x-6\right)\left(3x+4\right)\)

7: \(x^2+4x+4-49y^2\)

\(=\left(x^2+4x+4\right)-49y^2\)

\(=\left(x+2\right)^2-49y^2\)

=(x+2+7y)(x+2-7y)

8: \(a^3+9a-ab^2-6a^2\)

\(=a\left(a^2-6a+9-b^2\right)\)

\(=a\left[\left(a-3\right)^2-b^2\right]\)

\(=a\left(a-3-b\right)\left(a-3+b\right)\)

9: \(8x^2-16x+8-32y^2\)

\(=8\left(x^2-2x+1-4y^2\right)\)

\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)

=8(x-1-2y)(x-1+2y)

10: \(4x^2-4x+1-81a^2\)

\(=\left(4x^2-4x+1\right)-81a^2\)

\(=\left(2x-1\right)^2-\left(9a\right)^2\)

=(2x-1-9a)(2x-1+9a)

11: \(x^2-6xy+9y^2-121\)

\(=\left(x^2-6xy+9y^2\right)-121\)

\(=\left(x-3y\right)^2-11^2=\left(x-3y-11\right)\left(x-3y+11\right)\)

12: \(12x^2-24x+12-3y^2\)

\(=3\left(4x^2-8x+4-y^2\right)\)

\(=3\left[\left(2x-2\right)^2-y^2\right]=3\left(2x-2-y\right)\left(2x-2+y\right)\)

\(P=\dfrac{\left(x-1\right)^2+3}{\left(x-1\right)^2+5}=\dfrac{\left(x-1\right)^2+5-2}{\left(x-1\right)^2+5}=1-\dfrac{2}{\left(x-1\right)^2+5}\)

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

=>\(\dfrac{2}{\left(x-1\right)^2+5}< =\dfrac{2}{5}\forall x\)

=>\(-\dfrac{2}{\left(x-1\right)^2+5}>=-\dfrac{2}{5}\forall x\)

=>\(P=\dfrac{-2}{\left(x-1\right)^2+5}+1>=-\dfrac{2}{5}+1=\dfrac{3}{5}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(Q=\dfrac{\left(2y+3\right)^2-3}{\left(2y+3\right)^2+4}=\dfrac{\left(2y+3\right)^2+4-7}{\left(2y+3\right)^2+4}=1-\dfrac{7}{\left(2y+3\right)^2+4}\)

\(\left(2y+3\right)^2+4>=4\forall y\)

=>\(\dfrac{7}{\left(2y+3\right)^2+4}< =\dfrac{7}{4}\forall y\)

=>\(-\dfrac{7}{\left(2y+3\right)^2+4}>=-\dfrac{7}{4}\forall y\)

=>\(Q=-\dfrac{7}{\left(2y+3\right)^2+4}+1>=-\dfrac{3}{4}\forall y\)

Dấu '=' xảy ra khi 2y+3=0

=>2y=-3

=>y=-3/2

\(F=\dfrac{\left(x-1\right)^2+5}{\left(x-1\right)^2+2}=\dfrac{\left(x-1\right)^2+2+3}{\left(x-1\right)^2+2}=1+\dfrac{3}{\left(x-1\right)^2+2}\)

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

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

=>\(F=\dfrac{3}{\left(x-1\right)^2+2}+1< =\dfrac{5}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\\ =1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)-\dfrac{1}{x+1}\\ =1-\dfrac{1}{x+1}\\ =\dfrac{x+1}{x+1}-\dfrac{1}{x+1}\\ =\dfrac{x}{x+1}\)

\(5^{2x-1}=125\\ =>5^{2x-1}=5^3\\ =>2x-1=3\\ =>2x=3+1\\ =>2x=4\\ =>x=\dfrac{4}{2}\\ =>x=2\)

Vậy: ... 

5^{2x - 1} = 125

5^{2x - 1} = 5³

2x - 1 = 3

2x      = 3 +1

2x      = 4

  x      = 4 :2

  x      = 2

a, 19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26

=  ( 19 + 21 ) + ( 18 + 22 ) + ( 17 + 23 ) + ( 16 + 24 ) + ( 14 + 26 )

=   40 + 40 + 40 + 40 + 40

=   40 x 5

=    200

 Chúc bé học tốt

a.

19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26

= (19 + 21) + (18 + 22) + (17 + 23) + ( 16 + 24) + (14 + 26)

= 40 + 40 + 40 + 40 + 40

= 40 x 5

= 200

câu b mk đã lm ở dưới r nhé

c) 64 x 4 + 18 x 4 + 9 x 8

= 64 x 4 + 18 x 4 + 9 x 2 x 4

= 4 x ( 64 + 18 + 9 x 2)

=4 x (82 + 18)

= 4 x 100

= 400

\(15,2-x+12=93\)

\(15,2-x=93-12\)

\(15,2-x=81\)

\(x=15,2-81\)

\(x=-65,8\)

Em có nhầm ko nhỉ, lớp 4 sao đã có số âm được

lớp 4 chưa học số thập phân và số âm, hình như em viết sai đề bài.

x.16 - x.4 - x = 2

x. (16 - 4 - 1) = 2

x.       11        = 2

x                    = 2. 11

                  x = 22

           Vậy x = 22

x.16 - x.4 - x = 2

x.(16-4-1) = 2

x.11 = 2

x = 2:11

x = \(\dfrac{2}{11}\)

3 x 9 + 18 x 2 + 2 x 9 + 9 

= 3 x  9 + 9 x 2 x 2 + 2 x 9 + 9

= 9 x (3 + 2 + 2 + 2 + 1)

= 9 x 10

= 90

\(=3\times9+9\times2\times2+2\times9+9\)

\(=3\times9+4\times9+2\times9+9\)

\(=9\times\left(3+4+2+1\right)\)

\(=9\times10\)

\(=90\)