Hepppppppppppppppppppp

 `\color{#4DA6E6}{N} \color{#668EDD}{o} \color{#8077D5} {t} \color{#995FCD}{a} \color{#CC2FBC}{r} \color{#E618B3}{i} \color{#FF00AB}{a} \color{#EB5099}{2} \color{#EA4094}{3} \color{#EA2F90}{4}`

Câu `1:`

`a) {(x - y = 0),(x + y = 2):}`

`<=> {(x - y = 0),(2x = 2):}`

`<=> {(x = y),(x = 1):}`

`<=> {(y = 1),(x = 1):}`

Vậy `(x; y) = (1; 1)`

`b) {(3x - 2y = 9),(2x + 3y = -7):}`

`<=> {(6x - 4y = 18),(6x + 9y = -21):}`

`<=> {(3x - 2y = 9),(-13y = 39):}`

`<=> {(x =1),(y = -3):}`

Câu `2:`

Điều kiện xác định: `x^2 - 1 \ne 0`

`<=> (x - 1)(x + 1) \ne 0`

`<=> x \ne +-1`