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

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

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

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

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

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

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

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)

Băng Băng 2k6 giúp vs

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\left|3y+1\right|\ge0\forall y\)

Do đó: \(\left|2x-5\right|+\left|3y+1\right|\ge0\forall x,y\)

mà \(\left|2x-5\right|+\left|3y+1\right|=0\)

nên \(\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{-1}{3}\end{matrix}\right.\)

Vậy: \(x=\frac{5}{2}\) và \(y=\frac{-1}{3}\)

b) Ta có: \(\left|3x-4\right|\ge0\forall x\)

\(\left|3y-5\right|\ge0\forall y\)

Do đó: \(\left|3x-4\right|+\left|3y-5\right|\ge0\forall x,y\)

mà \(\left|3x-4\right|+\left|3y-5\right|=0\)

nên \(\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{5}{3}\end{matrix}\right.\)

Vậy: \(x=\frac{4}{3}\) và \(y=\frac{5}{3}\)

c) Ta có: |16-|x||≥0∀x

\(\left|5y-2\right|\ge0\forall y\)

Do đó: |16-|x||+|5y-2|≥0∀x,y

mà |16-|x||+|5y-2|=0

nên \(\left\{{}\begin{matrix}\text{|16-|x||}=0\\\left|5y-2\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16-\left|x\right|=0\\5y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|=16\\5y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{16;-16\right\}\\y=\frac{2}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{16;-16\right\}\) và \(y=\frac{2}{5}\)

có |2x-5| luôn \(\ge0\forall x\in Q\)

cũng có \(\left|3y+1\right|\ge0\forall y\in Q\)

=> \(\left|2x-5\right|+\left|3y-1\right|\ge0\forall x;y\in Q\)

=>\(\hept{\begin{cases}2x-5=0\\3y-1=0\end{cases}}\)<=> \(\hept{\begin{cases}2x=5\\3y=1\end{cases}}\)<=> \(\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{3}\end{cases}}\) 

vậy \(x=\frac{2}{5};y=\frac{1}{3}\)

em nhớ là phải dùng ngoặc nhọn như trên nhé! Nếu không sẽ sai đấy!

3 câu còn lại cũng tương tự

giúp mik câu cuối với các bạn

với câu cuối ;Nguyễn Khánh Linh  em chỉ cần tìm x ;  biến đổi vế rồi lắp x vào để giải tiếp

khúc đầu tương tự bài đầu

=> \(\hept{\begin{cases}2x-5=0\\xy-3y+2=0\end{cases}}\)<=> \(\hept{\begin{cases}x=\frac{5}{2}\\y\left(x-3\right)+2=0\end{cases}}\)<=> \(\hept{\begin{cases}x=\frac{5}{2}\\y\left(\frac{2}{5}-3\right)+2=0\end{cases}}\)

em tự giải tiếp

Lời giải:

Từ PT (2) suy ra $x=3y+1$

Từ PT (1) suy ra \(\left[{}\begin{matrix}2x+3y-2=0\\x-5y-3=0\end{matrix}\right.\)

Nếu $2x+3y-2=0$. Thay $x=3y+1$ vô thì:

$2(3y+1)+3y-2=0$

$\Leftrightarrow 9y=0\Leftrightarrow y=0$.

$x=3y+1=3.0+1=1$. HPT có nghiệm $(x,y)=(1,0)$

Nếu $x-5y-3=0$. Thay $x=3y+1$ vô thì:

$3y+1-5y-3=0$

$\Leftrightarrow -2y-2=0\Leftrightarrow y=-1$

$x=3(-1)+1=-2$. HPT có nghiệm $(x,y)=(-2; -1)$

\(\left(\frac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\le0\)(1)

Vì \(\left(\frac{1}{3}-2x\right)^{2018}\ge0\forall x\); \(\left(3y-x\right)^{2020}\ge0\forall x,y\)

\(\Rightarrow\left(\frac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\ge0\forall x,y\)(2)

Từ (1), (2) \(\Rightarrow\left(\frac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{3}-2x=0\\3y-x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=\frac{1}{18}\end{cases}}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}=6+18=24\left(đpcm\right)\)