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 = \(\dfrac{2}{11}\)

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

Vậy: ... 

5^{2x - 1} = 125

5^{2x - 1} = 5³

2x - 1 = 3

2x      = 3 +1

2x      = 4

  x      = 4 :2

  x      = 2

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

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

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

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

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

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

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

Gọi tổng số học sinh giỏi lớp 6;7;8 là x

Do x chia 3 dư 2, chia 4 dư 3, chia 53 dư 52

\(\Rightarrow x+1\) chia hết cho 3,4,53

\(\Rightarrow x+1\in BC\left(3;4;53\right)\)

Mà x nhỏ nhất 

\(\Rightarrow x+1=BCNN\left(3;4;53\right)\)

\(\Rightarrow x+1=636\)

\(\Rightarrow x=635\)

mọi người giúp e với bây giờ e đang gấp ạ!

                a; Giải:

Gọi phân số thứ nhất là \(\dfrac{a}{b}\) thì phân số thứ hai là:

       \(\dfrac{2}{5}\) - \(\dfrac{a}{b}\) 

Theo bài ra ta có: \(\dfrac{a}{b}\) - (\(\dfrac{2}{5}\) - \(\dfrac{a}{b}\)) = - \(\dfrac{4}{5}\)

                              \(\dfrac{a}{b}\) - \(\dfrac{2}{5}\) + \(\dfrac{a}{b}\) = - \(\dfrac{4}{5}\)

                              (\(\dfrac{a}{b}\) + \(\dfrac{a}{b}\)) = - \(\dfrac{4}{5}\) + \(\dfrac{2}{5}\)

                               2.\(\dfrac{a}{b}\) = - \(\dfrac{2}{5}\)

                               \(\dfrac{a}{b}\) = - \(\dfrac{2}{5}\) : 2

                                 \(\dfrac{a}{b}\) = - \(\dfrac{1}{5}\)

Phân số thứ hai là: \(\dfrac{2}{5}\)  - (- \(\dfrac{1}{5}\)) = \(\dfrac{3}{5}\) 

Kết luận:...

          b; Giải:

Gọi phân số thứ nhất là: \(\dfrac{a}{b}\) 

Phân số thứ hai là: \(\dfrac{12}{5}\) - \(\dfrac{a}{b}\)

Theo bài ra ta có:

    \(\dfrac{a}{b}\) : (\(\dfrac{12}{5}\) - \(\dfrac{a}{b}\)) = \(\dfrac{3}{7}\)

    \(\dfrac{a}{b}\) = \(\dfrac{3}{7}\) x (\(\dfrac{12}{5}\) - \(\dfrac{a}{b}\))

     \(\dfrac{a}{b}\) = \(\dfrac{36}{35}\)  - \(\dfrac{3}{7}\) x \(\dfrac{a}{b}\)

     \(\dfrac{a}{b}\) + \(\dfrac{3}{7}\) x \(\dfrac{a}{b}\) = \(\dfrac{36}{35}\)

     \(\dfrac{a}{b}\) x (1 + \(\dfrac{3}{7}\)) = \(\dfrac{36}{35}\)

      \(\dfrac{a}{b}\) x \(\dfrac{10}{7}\) = \(\dfrac{36}{35}\)

      \(\dfrac{a}{b}\)         = \(\dfrac{36}{35}\) : \(\dfrac{10}{7}\)

       \(\dfrac{a}{b}\)        = \(\dfrac{18}{25}\)

Phân số thứ hai là: 

      \(\dfrac{12}{5}\)  - \(\dfrac{18}{25}\) = \(\dfrac{42}{25}\)

Kết luận:...

giúp với...

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

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

\(\Rightarrow2x-2=2\)

\(\Rightarrow2x=2+2\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4\div2\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

`1/5 . (x+2)^2 + 1/3 . (2x - 2)^3 = 1/5 . (x+2)^2 + 1/3 . 2^3`

`<=> 1/5 . (x+2)^2 -  1/5 . (x+2)^2+ 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 0 + 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 1/3 : 1/3 . (2x - 2)^3 = 2^3`

`<=> 1 . (2x - 2)^3 = 2^3`

`<=> (2x - 2)^3 = 2^3`

`<=> 2x - 2 = 2`

`<=> 2x = 2+2 `

`<=> 2x = 4`

`<=> x = 4 : 2`

`<=> x = 2`

Vậy `x = 2`

\(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=4+5+\dfrac{3}{8}+\dfrac{2}{3}\)

\(=9+\dfrac{9}{24}+\dfrac{16}{24}\)

\(=9+\dfrac{25}{24}\)

\(=10\dfrac{1}{24}\)

\(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=\dfrac{35}{8}+\dfrac{17}{3}\)

\(=\dfrac{105}{24}+\dfrac{136}{24}\)

\(=\dfrac{241}{24}\)

\(\left(x^2+1\right)\left(x-1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-1=0

=>x=1