Đáp án:
Giải thích các bước giải:
Bài 1:
`a) B = 1 + 1/( x + 2 )` ( `Đk: x` $\neq$ `-2` )
Thay `x = ( -1 )/3 ( tmđk )` vào B , ta có:
`B = 1 + 1/( x + 2 )`
`<=> B = 1 + 1/( ( -1 )/3 + 2 )`
`<=> B = 1 + 1/( ( -1 )/3 + 6/3 )`
`<=> B = 1 + 1/( 5/3 )`
`<=> B = 1 + 3/5`
`<=> B = 8/5`
`b) A = 21/( x^2 - 9 ) - ( x - 4 )/( 3 - x ) - ( x - 1 )/( x + 3 ) `
`<=> A = 21/(( x - 3 )( x + 3 )) + (( x - 4 )( x + 3 ))/(( x - 3 )( x + 3 )) - (( x - 1 )( x - 3 ))/(( x - 3 )( x + 3 ))`
`Đk: x` $\neq$ `± 3`
`<=> A = 21/(( x - 3 )( x + 3 )) + ( x^2 - 4x + 3x - 12 )/(( x - 3 )( x + 3 )) - ( x^2 - 3x - x + 3 )/(( x - 3 )( x + 3 ))`
`<=> A = 21/(( x - 3 )( x + 3 )) + ( x^2 - x - 12 )/(( x - 3 )( x + 3 )) - ( x^2 - 4x + 3 )/(( x - 3 )( x + 3 ))`
`<=> A = ( 21 + x^2 - x - 12 - x^2 + 4x - 3 )/(( x - 3 )( x + 3 ))`
`<=> A = ( 3x + 6 )/(( x - 3 )( x + 3 ))`
`<=> A = ( 3( x + 2 ))/(( x - 3 )( x + 3 ))`
`c) C = A.B`
`<=> C = ( 3( x + 2 ))/(( x - 3 )( x + 3 )) . ( 1 + 1/( x + 2 ) )`
`<=> C = ( 3( x + 2 ))/(( x - 3 )( x + 3 )) . ( x + 2 + 1 )/( x + 2 )`
`<=> C = ( 3( x + 2 ))/(( x - 3 )( x + 3 )) . ( x + 3 )/( x + 2 )`
`<=> C = 3/( x - 3 )`
`C > 0 <=> 3/( x - 3 ) > 0`
`<=> x -3 > 0` ( vì `3 > 0` )
`<=> x > 3`
Vậy `x > 3` thì `C > 0`
