Ta có: ||3x-3|+2x+\(\left(-1\right)^{2016}\) |=3x+\(2017^0\)

=>||3x-3|+2x+1|=3x+1(1)

TH1: |3x-3|+2x+1=3x+1

=>|3x-3|=3x+1-2x-1=x

=>\(\begin{cases}x\ge0\\ x^2=\left(3x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(3x-3-x\right)\left(3x-3+x\right)=0\end{cases}\)

=>\(\begin{cases}x\ge0\\ \left(2x-3\right)\left(4x-3\right)=0\end{cases}\Rightarrow x\in\left\lbrace\frac32;\frac34\right\rbrace\)

THay lại vào trong (1), ta thấy cả x=3/2 và x=3/4 đều thỏa mãn

TH2: |3x-3|+2x+1=-3x-1

=>|3x-3|=-3x-1-2x-1

=>|3x-3|=-5x-2

=>\(\begin{cases}-5x-2\ge0\\ \left(-5x-2\right)^2=\left(3x-3\right)^2\end{cases}\Rightarrow\begin{cases}-5x\ge2\\ \left(5x+2\right)^2-\left(3x-3\right)^2=0\end{cases}\)

=>\(\begin{cases}x\le-\frac25\\ \left(5x+2-3x+3\right)\left(5x+2+3x-3\right)=0\end{cases}\Rightarrow\begin{cases}x\le-\frac25\\ \left(2x+5\right)\left(8x-1\right)=0\end{cases}\)

=>\(x=-\frac52\)

Thay lại vào (1), ta thấy x=-5/2 không thỏa mãn

=>Loại

Vậy: \(x\in\left\lbrace\frac32;\frac34\right\rbrace\)

x=0

ban

a, (2x-3)⁴=(2x-3)⁶

=> (2x-3)⁶ : (2x-3)⁴=1

=> (2x-3)³=

=> 2x-3=1

=> 2x=4

=> x=2

b, (3x+5)³=(3x+5)²⁰¹⁶

=> (3x+5)²⁰¹⁶ : (3x+5)³=1

=> (3x+5)²⁰¹³=1

=> 3x+5=1

=> 3x=-4

=> x=-4/3

c, (2x+1)²⁰¹⁵=(2x+1)²⁰¹⁷

=> (2x+1)²⁰¹⁷ : (2x+1)²⁰¹⁵=1

=> (2x+1)²=1

=> 2x+1=1

=> 2x=0

=> x=0

\(a)\) \(\left|\left|3x-3\right|2x+\left(-1\right)^{2016}\right|=3x+2017^0\)

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

Mà \(\left|\left|3x-3\right|2x+1\right|\ge0\) nên \(3x+1\ge0\)\(\Rightarrow\)\(x\ge1\)

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

\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3x}{2x}\)

\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3}{2}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-3=\frac{3}{2}\\3x-3=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{9}{2}\\3x=\frac{3}{2}\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{9}{2}:3\\x=\frac{3}{2}:3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(tmx\ge1\right)\\x=\frac{1}{2}\left(loai\right)\end{cases}}}\)

Vậy \(x=\frac{3}{2}\)

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