Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

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

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

Ta có \(10\left(x-1\right)^{20}+20\left(y+2\right)^{10}=0\)

=> \(\hept{\begin{cases}10\left(x-1\right)^{20}=0\\20\left(y+2\right)^{10}=0\end{cases}}\)=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\)=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Thay x = 1 và y = -2 vào biểu thức \(2x^3+3y^2-14\), ta có:

\(2.1^3+3\left(-2\right)^2-14=2+12-14=0\)

Vậy giá trị của biểu thức \(2x^3+3y^2-14\)là 0 khi \(10\left(x-1\right)^{20}+20\left(y+2\right)^{10}=0\).

\(\left|2x-27\right|^{2017}+\left(3y+27\right)^{2016}=0\)

\(\Rightarrow\left|2x-27\right|^{2017}=0\) và \(\left(3y+27\right)^{2016}=0\)

+) \(\left|2x-27\right|^{2017}=0\Rightarrow2x-27=0\Rightarrow2x=27\Rightarrow x=\frac{27}{2}\)

+) \(\left(3y+27\right)^{2016}=0\Rightarrow3y+27=0\Rightarrow3y=-27\Rightarrow y=-9\)

Vậy \(x=\frac{27}{2};y=-9\)

ta có:

|2x-27|²⁰¹⁷≥0

(3y+27)²⁰¹⁶ ≥0

vậy |2x-27|²⁰¹⁷+(3y+37)²⁰¹⁶ ≥0

dấu "=" xảy ra khi

|2x-27|²⁰¹⁷=(3y+27)²⁰¹⁶=0

|2x-27|²⁰¹⁷=0

=> 2x=27

=>x=27/2

(3y+27)²⁰¹⁶=0

=> 3y=-27

=> y=-9

vậy với x=27/2 và y=-9 thì x,y thỏa mãn yêu cầu đề bài

1,

Vì \(\left|2x-27\right|^{2007}\ge0;\left(3y+10\right)^{2008}\ge0\)

\(\Rightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)

Mà \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)

\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)

2,

TH1: \(x\ge\frac{3}{5}\)

<=> 2(5x-3)-2x=14

<=> 10x-6-2x=14

<=>8x-6=14

<=>8x=20

<=>x=5/2 (thỏa mãn)

TH2: x < 3/5

<=> 2(3-5x)-2x=14

<=>6-10x-2x=14

<=>6-12x=14

<=>12x=-8

<=>x=-2/3 (thỏa mãn)

Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)

1 x=13,5 ;y=-10/3

2 kết quả x =-2/3

ko bieet

Ta có:\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)+15xy\left(y-x\right)+1\)Thế \(x-y=0\) vào C ta được:

\(C=0+0+0+1\)

C = 0

A=2(x+y)+3xy(x+y)+5x²y²(x+y)+2

A=2.0+3xy.0+5x²y².0+2

A=2

B=xy(x+y)+2x²y (x+y)+5

B=xy.0+2x²y.0+5=5

a,Ta có 2(x+y)+3xy(x+y)+5x²y²(x+y)+4

Xg thay x+y=0 vào là dc bn nhó

Chúc bn hok tốt

\(A=\left|x-3\right|+\left|y+3\right|+2016\)

\(\left|x-3\right|\ge0\)

\(\left|y+3\right|\ge0\)

\(\Rightarrow\left|x-3\right|+\left|y+3\right|+2016\ge2016\)

Dấu ''='' xảy ra khi \(x-3=y+3=0\)

\(x=3;y=-3\)

\(MinA=2016\Leftrightarrow x=3;y=-3\)

\(\left(x-10\right)+\left(2x-6\right)=8\)

\(x-10+2x-6=8\)

\(3x=8+10+6\)

\(3x=24\)

\(x=\frac{24}{3}\)

x = 8

Ta có:

\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1\)

=\(0+0+0+1=1\)

\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)

\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)\)

\(=0+0+1=1\)

~^~