Lời giải:

Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)

\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)

Do đó:

\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)

\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)

............

\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)

Cộng theo vế:

\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)

\(=\underbrace{1+1+1...+1}_{1008}=1008\)

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)

\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)

\(B=2x^2+2x+2012^2+2013^2\)

\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)

\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)

\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)

Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)

đầu bài trên tớ làm luôn nhá !!!

a,  / 3x+1/= 5-3

    / 3x+1/= 2

   3x+1=2

  x+1 = 2:3 

 x+1 = 2 phần 3

x= 2/3 -1 

x= -1/3 

còn phần b.c.d lần lượt nha bạn 

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

( 3x - 2⁴ ) . 7⁵ = 2.7⁶ .1/2017⁰

( 3x - 2⁴ ) . 7⁵ =235298

( 3x - 2⁴ ) = 235298 : 7⁵ 

( 3x - 2⁴ ) =14

3x = 14 + 2⁴ 

3x = 30

x = 0

dung 100%

\(\left(3x-2^4\right).7^5=2.7^6.\frac{1}{2017^0}\)

\(\Leftrightarrow\left(3x-16\right).7^5=2.7^6.1\)

\(\Leftrightarrow3x-16=\frac{2.7^6}{7^5}\)

\(\Leftrightarrow3x-16=2.7\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\)

Nhưng mình làm cách này có được không :  (Tự nhiên nghĩ được)

(3x-2^4).7^5=2.7^6.1/2017^0 

(3x-2^4).7^5=2.7.7^5.1

3x-16=2.7   ( Bớt cả hai vế cho 7^5)

3x=14+16

x=30:3

x=10

x=1/3 hoặc x=2/3 bạn nhé.

pt <=> \(\left(3x-1\right)^{2013}-\left(3x-1\right)^{2015}=0\)

\(\Leftrightarrow\left(3x-1\right)^{2013}\left(\left(3x-1\right)^2-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)^{2013}\left(3x-2\right)3x=0\)

\(\orbr{\begin{cases}x=0\\3x-1=0,3x-2=0\end{cases}}\)

Vậy x=0, x=1/3,hoặc x=2/3

b)\(\left|21x-5\right|=\left|3x-7\right|\)

\(\Leftrightarrow\begin{cases}21x-5=3x-7\\21x-5=7-3x\end{cases}\)

\(\Leftrightarrow\begin{cases}9x=-1\\24x=12\end{cases}\)

\(\Leftrightarrow\begin{cases}x=-\frac{1}{9}\\x=\frac{1}{2}\end{cases}\)

a)\(\left|2x-7\right|=3\)

\(\Rightarrow2x-7=\pm3\)

Nếu \(2x-7=3\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

Nếu \(2x-7=-3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

c)\(\left|7-2x\right|=1-3x\)

\(\Rightarrow7-2x=\pm\left(1-3x\right)\)

Nếu \(7-2x=1-3x\)

\(\Rightarrow x=-6\)

Nếu \(7-2x=-\left(1-3x\right)\)

\(\Rightarrow7-2x=-1+3x\)

\(\Rightarrow5x=8\)

\(\Rightarrow x=\frac{8}{5}\)