Question

1)Tìm x thuộc Z:

a)(x-2)²-9=7

b)/x-2/-9=7

2) Tìm x,y thuộc Z:

a)/x-5/+/y-7/≤0

b)/x+3/+(y+2019)²≤0

Answer 1

1a) (x - 2)² - 9 = 7

=> (x - 2)² = 7 + 9

=> (x - 2)² = 16

=> (x - 2)² = 4²

=> \(\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy ...

Answer 2

1b) |x - 2| - 9 = 7

=> |x - 2| = 7 + 9

=> |x - 2| = 16

=> \(\orbr{\begin{cases}x-2=16\\x-2=-16\end{cases}}\)

=> \(\orbr{\begin{cases}x=18\\x=-14\end{cases}}\)

Answer 3

2a) |x - 5| + |x - 7| \(\le\)0

Ta có: |x - 5| \(\ge\)0 \(\forall\)x

          |y - 7| \(\ge\) 0 \(\forall\)y

=> |x - 5| + |y - 7| \(\ge\)0 \(\forall\)x,y

+) Với |x - 5| + |y - 7| = 0

=> \(\hept{\begin{cases}x-5=0\\y-7=0\end{cases}}\)

=>  \(\hept{\begin{cases}x=5\\y=7\end{cases}}\)

+) Với |x - 5| + |y - 7| < 0

=> ko có giá trị x,y nào thõa mãn 

Answer 4

\(\left(x-2\right)^2-9=7\)

\(\Rightarrow\left(x-2\right)^2=16=\left(\pm4\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}}\)

\(\left|x-2\right|-9=7\)

\(\Leftrightarrow\left|x-2\right|=16\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=16\\x-2=-16\end{cases}\Leftrightarrow\orbr{\begin{cases}x=18\\x=-14\end{cases}}}\)

a;\(\left|x-5\right|+\left|y-7\right|\le0^{\left(1\right)}\)

Ta có \(\left|x-5\right|\ge0\forall x;\left|y-7\right|\ge0\forall y\)

\(\Rightarrow\left|x-5\right|+\left|y-7\right|\ge0\forall x;y\)

\(\Rightarrow\left|x-5\right|+\left|y-7\right|\ge0^{\left(2\right)}\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\left|x-5\right|+\left|y-7\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\y-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=7\end{cases}}}\)

Vậy....................................

b;\(\left|x+3\right|+\left(y+2019\right)^2\le0^{\left(1\right)}\)

Ta có \(\left|x+3\right|\ge0\forall x;\left(y+2019\right)^2\ge0\forall y\)

\(\Rightarrow\left|x+3\right|+\left(y+2019\right)^2\ge0\forall x;y\)

\(\Rightarrow\left|x+3\right|+\left(y+2019\right)^2\ge0^{\left(2\right)}\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\left|x+3\right|+\left(y+2019\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\y+2019=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-2019\end{cases}}}\)

Vậy.............................