\(.a.\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}=0\)

\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Leftrightarrow\left[\begin{matrix}\left(x-7\right)^{x+1}=0\\\left[1-\left(x-7\right)^{10}\right]=0\end{matrix}\right.\)

+ Nếu \(\left(x-7\right)^{x+1}=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=0+7\)

\(\Rightarrow x=7\)

+ Nếu \(1-\left(x-7\right)^{10}=0\)

\(\Rightarrow\left(x-7\right)^{10}=1\)

\(\Rightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)

\(\Rightarrow\left[\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=1+7\\x=-1+7\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Vậy : \(x\in\left\{6;7;8\right\}\)

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

(2x-5)^2008 > 0

(3y+4)^2010 > 0

=>(2x-5)^2008+(3y+4)^2010>0

mà theo đề:(2x-5)^2008+(3y+4)^2010 < 0

=>(2x-5)^2008=(3y+4)^2010=0

+)(2x-5)^2008=0=>2x=5=>x=5/2

+)(3y+4)^2010=0=>3y=-4=>y=-4/3

Vậy...

vì 2008và 2010 chẵn nên (2x-5)^2008 và(3y+4)^2010> hoac = 0Vậy=0

x=5/2 và y =-4/3

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...

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

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

2023 =))

d: =>6y+2-4x+4=5 và 15y+5-8x+8=9

=>-4x+6y=-1 và -8x+15y=-4

=>x=-3/4; y=-2/3

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

b: \(\Leftrightarrow\left\{{}\begin{matrix}3y-15+2x-6=0\\7x-28+3y+3y-3=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=21\\7x+6y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{19}{3}\end{matrix}\right.\)