mux 2 nha moi ng 

2016 : \(\left[25-\left(3x+2\right)\right]=3^7\)

<=> 25 - (3x + 2) = 2016 : 2187

<=> 25 - 3x - 2 = \(\dfrac{224}{243}\)

<=> -3x = \(\dfrac{224}{243}\) + 2 - 25

<=> -3x = \(\dfrac{-5365}{243}\)

<=> x = 7,359396433 \(\approx\) 7,4

2016 : \(\left[25-\left(3x+2\right)\right]=3^2\)

<=> 25 - (3x + 2) = 2016 : 9

<=> 25 - 3x - 2 = 224

<=> -3x = 224 - 25 + 2

<=> -3x = 201

<=> x = \(\dfrac{201}{-3}=-67\)

Ta có: \(2016:\left[25-\left(3x+2\right)\right]=32\cdot7\)

\(\Leftrightarrow25-\left(3x+2\right)=9\)

\(\Leftrightarrow3x+2=16\)

\(\Leftrightarrow3x=14\)

hay \(x=\dfrac{14}{3}\)

2016 : [25 - (3x+2)]=32.7

25 - (3x+2)= 2016 : 224

25 - (3x+2)= 9

3x+2 = 16

3x=14

x=\(\dfrac{14}{3}\)

2016 : [25 - (3x+2)]=32.7

25 - (3x+2)= 2016 : 224

25 - (3x+2)= 9

3x+2 = 16

3x=14

x=14/3

Đặt: \(\left\{{}\begin{matrix}a=2x^2+x-2016\\b=x^2-3x-1000\end{matrix}\right.\). Phương trình trở thành:

\(a^2+4b^2=4ab\)

\(\Leftrightarrow a^2-4ab+4b^2=0\)

\(\Leftrightarrow\left(a-2b\right)^2=0\Leftrightarrow a=2b\)

\(\Rightarrow2x^2+x-2016=2\left(x^2-3x-1000\right)\)

\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)

Vậy: \(x=\dfrac{16}{7}\)

\(\left(\frac{2}{3}x-1\right)\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)

=>\(\frac{2}{3}x-1=0\) hoặc \(\frac{3}{4}x+\frac{1}{2}=0\)

+)Nếu \(\frac{2}{3}x-1=0\)

=>\(\frac{2}{3}x=1\Rightarrow x=\frac{3}{2}\)

+)Nếu \(\frac{3}{4}x+\frac{1}{2}=0\)

=>\(\frac{3}{4}x=-\frac{1}{2}\Rightarrow x=-\frac{2}{3}\)

Vậy \(x=\frac{3}{2}\) hoặc \(x=-\frac{2}{3}\)

Còn nhỏ nên không hiểu chi hết *,.,*

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

\(=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2016\right)\left(\sqrt[3]{1+3x}-1\right)+x^2}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{3x\left(x^2+2016\right)}{\sqrt[3]{\left(1+3x\right)^2}+\sqrt[3]{1+3x}+1}+x^2}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{3\left(x^2+2016\right)}{\sqrt[3]{\left(1+3x\right)^2}+\sqrt[3]{1+3x}+1}+x\right)=\dfrac{3.2016}{3}=2016\)