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

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b) Ta có: \(\hept{\begin{cases}\left|2x-27\right|^{2007}\ge0\left(\forall x\right)\\\left(3y+10\right)^{2008}\ge0\left(\forall y\right)\end{cases}}\)

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

\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2017}=0\\\left(3y+10\right)^{2008}=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}=13,5\\y=-\frac{10}{3}\end{cases}}\)

Đẳng thức: \(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

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

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

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

Thay vào \(M=\left(x+y\right)^{2007}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\) ta được:

\(M=\left(1-1\right)^{2007}+\left(1-2\right)^{2008}+\left(-1+1\right)^{2009}=\left(-1\right)^{2008}=1\)

Ta có:

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow x^2+4x^2+y^2+4y^2+8xy-2x+2y+1+1=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)+\left(4x^2+8xy+4y^2\right)=0\)

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

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

Mà: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\\4\left(x+y\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2+4\left(x+y\right)^2\ge0\)

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

Thay giá trị x và y vào M ta có:

\(M=\left(x+y\right)^{2007}+\left(x-2\right)^{2008}+\left(y+1\right)^{2009}\)

\(M=\left(1-1\right)^{2007}+\left(1-2\right)^{2008}+\left(-1+1\right)^{2009}\)

\(M=0^{2007}+\left(-1\right)^{2008}+0^{2009}\)
\(M=\left(-1\right)^{2008}\)

\(M=1\)

Vì \(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\);  \(\left(3y+10\right)^{2012}\ge0\)

=>\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)

Dấu "=" xảy ra khi \(\left|2x-27\right|^{2011}=\left(3y+10\right)^{2012}=0\Leftrightarrow\hept{\begin{cases}\left|2x-27\right|=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)

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

Do đó: \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\) với mọi x.

Theo đề bài, ta có:

\(\left|2x-27\right|^{2007}=0\Rightarrow2x-27=0\Rightarrow x=....\)

\(\left(3y+10\right)^{2008}=0\Rightarrow3y+10=0\Rightarrow y=.....\)

x=27/2

y= -10/3

Vì /2x-27/^2007 > 0 với mọi x; (3y+10)^2008 > 0 với mọi x

Do đó:/2x-27/^2007 + (3y+10)^2008 > 0 với mọi x(mấy câu này mình thêm vào để bạn hiểu hơn thôi)

Theo đề bài thì ta có:/2x-27/^2007+(3y+10)^2008 =0

=>/2x-27/^2007 =0 =>2x-27=0 =>x=....

(3y+10)^2008 =0 =>3y+10=0 =>y=.....