Question

Tìm x,y:

a,/x-1/+/x-3/<x+1.

b,/x+y+2/+/2y+1/<hoặc=0.

c,/x-y-5/+(y-3)^2016=0.

Answer 1

a) |x - 1| + |x - 3| < x + 1

Có: \(\left|x-1\right|+\left|x-3\right|\ge\left|x-1+3-x\right|=\left|2\right|=2\)

=> x + 1 > 2

=> x > 1

+ Với x < 3 thì |x - 1| + |x - 3| = (x - 1) + (3 - x) = 2

Mà x + 1 > 1 + 1 = 2 do x > 1, thỏa mãn

+ Với \(x\ge3\) thì |x - 1| + |x - 3| = (x - 1) + (x - 3) = 2x - 4 < x + 1

=> 2x - x < 1 + 4

=> x < 5

Vậy \(\left[\begin{array}{nghiempt}1< x< 3\\3\le x< 5\end{array}\right.\) thỏa mãn đề bài

 

 

Answer 2

b) Có: \(\left|x+y+2\right|\ge0;\left|2y+1\right|\ge0\forall x;y\)

\(\Rightarrow\left|x+y+2\right|+\left|2y+1\right|\ge0\)

Mà theo đề bài: \(\left|x+y+2\right|+\left|2y+1\right|\le0\)

=> |x + y + 2| + |2y + 1| = 0

\(\Rightarrow\begin{cases}\left|x+y+2\right|=0\\\left|2y+1\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x+y+2=0\\2y+1=0\end{cases}\)\(\Rightarrow\begin{cases}x+y=-2\\2y=-1\end{cases}\)\(\Rightarrow\begin{cases}x+y=-2\\y=\frac{-1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{-3}{2}\\y=\frac{-1}{2}\end{cases}\)

Vậy \(x=\frac{-3}{2};y=\frac{-1}{2}\) thỏa mãn đề bài

Answer 3

b)Ta thấy:\(\begin{cases}\left|x+y+2\right|\ge0\\\left|2y+1\right|\ge0\end{cases}\)

\(\Rightarrow\left|x+y+2\right|+\left|2y+1\right|\ge0\)

Mà \(\left|x+y+2\right|+\left|2y+1\right|\le0\)

\(\Rightarrow\left|x+y+2\right|+\left|2y+1\right|=0\)

\(\Rightarrow\begin{cases}\left|x+y+2\right|=0\\\left|2y+1\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x+y+2=0\\2y+1=0\end{cases}\)

\(\Rightarrow\begin{cases}x+y+2=0\\y=-\frac{1}{2}\end{cases}\)\(\Rightarrow\begin{cases}x+\left(-\frac{1}{2}\right)+2=0\\y=-\frac{1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{3}{2}\\y=-\frac{1}{2}\end{cases}\)

 

Answer 4

c) Có: \(\left|x-y-5\right|\ge0;\left(y-3\right)^{2016}\ge0\forall x;y\)

Mà theo đề bài: |x - y - 5| + (y - 3)²⁰¹⁶ = 0

\(\Rightarrow\begin{cases}\left|x-y-5\right|=0\\\left(y-3\right)^{2016}=0\end{cases}\)\(\Rightarrow\begin{cases}x-y-5=0\\y-3=0\end{cases}\)\(\Rightarrow\begin{cases}x-y=5\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=8\\y=3\end{cases}\)

Vậy x = 8, y = 3 thỏa mãn đề bài

Answer 5

c)Ta thấy:\(\begin{cases}\left|x-y-5\right|\ge0\\\left(y-3\right)^{2016}\ge0\end{cases}\)

\(\Rightarrow\left|x-y-5\right|+\left(y-3\right)^{2016}\ge0\)

Mà \(\left|x-y-5\right|+\left(y-3\right)^{2016}=0\)

\(\Rightarrow\begin{cases}\left|x-y-5\right|=0\\\left(y-3\right)^{2016}=0\end{cases}\)\(\Rightarrow\begin{cases}x-y-5=0\\y-3=0\end{cases}\)

\(\Rightarrow\begin{cases}x-y-5=0\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x-3-5=0\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=8\\y=3\end{cases}\)