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

bài 2 : 

\(\left(3y-1\right)^{10}=\left(3y-1\right)^{20}\)

\(\Rightarrow\left(3y-1\right)^{20}-\left(3y-1\right)^{10}=0\)

\(\Rightarrow\left(3y-1\right)^{10}\left[\left(3y-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3y-1\right)^{10}=0\\\left(3y-1\right)^{10}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3y-1=0\\\left(3y-1\right)^{10}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3y=1\\3y-1=\pm1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=\frac{1}{3}\\y=0\text{ }or\text{ }y=\frac{2}{3}\end{cases}}\)

BÀI  3

\(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Rightarrow\left(x-5\right)^2-\left(1-3x\right)^2=0\)

\(\Rightarrow\left(x-5-1+3x\right)\left(x-5+1-3x\right)=0\)

\(\Rightarrow\left(4x-6\right)\left(-2x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4x-6=0\\-2x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}}\)

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 \(y=\frac{2}{5}\)

2023 =))

\(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)(1)

\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)(2)

Từ (1) và (2) suy ra \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-x+3}{4+9-4}=\frac{45}{9}=5\)

\(\Rightarrow\hept{\begin{cases}x=\left(5.4+2\right):2=11\\y=\left(5.9+6\right):3=17\\z=\left(4.5+3\right)=23\end{cases}}\)

a.  x=1      y= -3

b.  x=5      y=7/2

c.  x= -1    y= -1/2

d.  x=1/4   y= 1/4

a) x = 1    

y = -3

b) x = 5

y = 7/2

c) x = -1

y = -1/2

d) x = 1/4 

y = 1/4

nha bn