Ta có:

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

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

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

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

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

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

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

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

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

Vậy: ... 

5^{2x - 1} = 125

5^{2x - 1} = 5³

2x - 1 = 3

2x      = 3 +1

2x      = 4

  x      = 4 :2

  x      = 2

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 = \(\dfrac{2}{11}\)

A = {x l x ≤ 12 l x ϵ N}

A = {0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12}

C1:

\(A=\left\{0;1;2;3;4;5;6;7;8;9;10;11;12\right\}\)

\(C2:A=\left\{x\inℕ|x\le12\right\}\)

\(15)15+4\left(x-2\right)=95\\ 4\left(x-2\right)=95-15\\ 4\left(x-2\right)=80\\ x-2=\dfrac{80}{4}\\ x-2=20\\ x=20+2\\ x=22\\ 16)20-\left(x+14\right)=5\\ x+14=20-5\\ x+14=15\\ x=15-14\\ x=1\\ 17)24+3\left(5-x\right)=27\\ 3\left(5-x\right)=27-24\\ 3\left(5-x\right)=3\\ 5-x=\dfrac{3}{3}=1\\ x=5-1=4\\ 18)15:x=5\\ x=15:5\\ x=3\\ 19)\dfrac{x}{4}=3\\ x=4\cdot3\\ x=12\\ 20)\dfrac{21}{x}=7\\ x=\dfrac{21}{7}\\ x=3\)

`23*16+23*84-300`

`=23*(16+84)-300`

`=23*100-300`

`=2300-300`

`=2000 `

Gọi d là ƯCLN của `2n+3` và `4n+7`

Ta có: 

`2n+3` ⋮ d và `4n+7` ⋮ d 

`=>2(2n+3)` ⋮ d và `4n+7` ⋮ d

`=>4n+6` ⋮ d và `4n+7` ⋮ d 

`=>(4n+7)-(4n+6)` ⋮ d 

`=>1` ⋮ d 

`=>d=1` 

Vậy: `2n+3` và `4n+7` là 2 số nguyên tố cùng nhau 

Mọi người giúp em với ạ, bây giờ em đang cần gấp ạ!

a: Các góc có trong hình vẽ là \(\widehat{tMN};\widehat{zMN};\widehat{tMz}\)

b: Góc bẹt là \(\widehat{tMz}\)