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 : \(\Leftrightarrow x=-\dfrac{1}{2}\)

a.\(\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)

\(x-\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)

\(x-\frac{1}{3}=\frac{1}{3}\)

\(x=\frac{2}{3}\)

\(a.\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\frac{5}{6}-x=\frac{1}{6}\)

\(\Leftrightarrow\frac{5}{6}-\frac{1}{6}=x\)

\(\Leftrightarrow x=\frac{2}{3}\)

\(b.||3x+2|-2x-5|=3x-\left(-1\right)^{2015}\)

\(\Leftrightarrow||3x+2|-2x-5|=3x+1\)

\(\Leftrightarrow\orbr{\begin{cases}|3x+2|-2x-5=3x+1\\|3x+2|-2x-5=-3x-1\end{cases}\Leftrightarrow\orbr{\begin{cases}|3x+2|=5x+6\left(n\right)\\|3x+2|=-\left(x-4\right)\left(l\right)\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=5x+6\\3x+2=-5x-6\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x=4\\8x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)

V...\(x=-1;x=-2\)

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 

a/ \(\left(4x-5\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy ............

b/ \(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x+2}{2015}+1\right)=\left(\dfrac{x+3}{2014}+1\right)+\left(\dfrac{x+4}{2013}+1\right)\)

\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)

\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)

\(\Leftrightarrow x+2017\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

Mà \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)

\(\Leftrightarrow x+2017=0\)

\(\Leftrightarrow x=-2017\)

Vậy ..

\(\left(4x-5\right)\left(3x+2\right)=0\)

\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)

\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)

\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)

\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)

\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)

Nên:

\(x+2017=0\Rightarrow x=-2017\)

(3x - 7)²⁰¹⁵ = (3x - 7)²⁰¹⁷

(3x - 7)²⁰¹⁷ - (3x - 7)²⁰¹⁵ = 0

(3x - 7)²⁰¹⁷[(3x - 7)² - 1] = 0

=> (3x - 7)²⁰¹⁷ = 0 hoặc (3x - 7)² = 1

=> 3x - 7 = 0 hoặc 3x - 7 = ± 1

=> x = 7/3 hoặc x = { 8/3 ; 2 }

Vậy x = { 2; 7/3; 8/3 }

\(y\left(y^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y^2-1=0\end{cases}}\)

(3x - 7)^2015 = (3x - 7)^2017

(3x - 7)^2017 - (3x - 7)^2015 = 0

(3x - 7)^2017 [(3x - 7)^2 - 1] = 0

=> (3x - 7)^2017 = 0 hoặc (3x - 7)^2 = 1

=> 3x - 7 = 0 hoặc 3x - 7 = ± 1

=> x = 7/3 hoặc x = { 8/3 ; 2 }

Vậy x = { 2; 7/3; 8/3 }