9.5 đ

−5 . 29 −12 . −10 8 . 5 29 = ( − 5 ) . ( − 12 ) . 8.29 8.29. ( − 10 ) . 5 = 8.29.(−10).5 (−5 ).(−12).8.29 = − 5 . 12.8.29 8.29.10.5 =− 8.29.10.5 5 .12.8.29 = − 12 10 =− 10 12 = − 6 5 =− 5 6

5 . 29 −12 . −10 8 . 5 29 = ( − 5 ) . ( − 12 ) . 8.29 8.29. ( − 10 ) . 5 = 8.29.(−10).5 (−5 ).(−12).8.29 = − 5 . 12.8.29 8.29.10.5 =− 8.29.10.5 5 .12.8.29 = − 12 10 =− 10 12 = − 6 5 =− 5 6

=> ( 2 x − 1 ) ( y − 1 2 ) = 5 2 (2x−1)(y− 2 1 )= 2 5 =>(2x-1)(2y-1)=5 => ( 2 x − 1 ; 2 y − 1 ) ∈ { ( 1 ; 5 ) ; ( 5 ; 1 ) ; ( − 1 ; − 5 ) ; ( − 5 ; − 1 ) } (2x−1;2y−1)∈{(1;5);(5;1);(−1;−5);(−5;−1)} => ( x ; y ) ∈ { ( 1 ; 3 ) ; ( 3 ; 1 ) ; ( 0 ; − 2 ) ; ( − 2 ; 0 ) } (x;y)∈{(1;3);(3;1);(0;−2);(−2;0)}

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Kiến thức cần nhớ: Muốn chuyển số thập phân sang phân số ta cần thực hiện các bước sau: Bước 1: biểu diễn số thập phân dưới dạng phân số thập phân. Bước 2: Rút gọn phân số thập phân trên ta được phân số tối giản thỏa mãn yêu cầu Giải: - 3,2 = - 32 10 10 32 = - 16 5 5 16

: 1 + 1 2 ( 1 + 2 ) + 1 3 ( 1 + 2 + 3 ) + . . . + 1 200 ( 1 + 2 + . . . + 200 ) 1+ 2 1 (1+2)+ 3 1 (1+2+3)+...+ 200 1 (1+2+...+200) = 1 + 1 2 ⋅ 2 ⋅ 3 2 + 1 3 ⋅ 3 ⋅ 4 2 + . . . + 1 200 ⋅ 200 ⋅ 201 2 =1+ 2 1 ⋅ 2 2⋅3 + 3 1 ⋅ 2 3⋅4 +...+ 200 1 ⋅ 2 200⋅201 = 1 + 3 2 + 4 2 + . . . + 201 2 =1+ 2 3 + 2 4 +...+ 2 201 = 2 + 3 + . . . + 201 2 = ( 201 − 2 + 1 ) ⋅ ( 201 + 2 ) 2 2 = 2 2+3+...+201 = 2 (201−2+1)⋅ 2 (201+2) = 200 ⋅ 203 4 = 50 ⋅ 203 = 10150 = 4 200⋅203 =50⋅203=10150 d: 2 7 ⋅ 3 + 2 10 13 ⋅ 2 7 − 14 ⋅ 2 6 = 2 7 ( 3 + 2 3 ) 2 6 ( 13 ⋅ 2 − 14 ) = 2 ⋅ 3 + 8 26 − 14 13⋅2 7 −14⋅2 6 2 7 ⋅3+2 10 = 2 6 (13⋅2−14) 2 7 (3+2 3 ) =2⋅ 26−14 3+8 = 2 ⋅ 11 12 = 11 6 =2⋅ 12 11 = 6 11 Bài 2: 1: a: ( x − 1 2 ) : 2 3 + 0 , 75 = − 3 3 4 (x− 2 1 ): 3 2 +0,75=−3 4 3 => ( x − 1 2 ) : 2 3 = − 3 − 3 4 − 3 4 = − 3 − 3 2 = − 9 2 (x− 2 1 ): 3 2 =−3− 4 3 − 4 3 =−3− 2 3 =− 2 9 => x − 1 2 = − 9 2 ⋅ 2 3 = − 9 3 = − 3 x− 2 1 =− 2 9 ⋅ 3 2 =− 3 9 =−3 => x = − 3 + 1 2 = − 5 2 x=−3+ 2 1 =− 2 5 b: ( 2 x − 15 ) 3 = ( 2 2 ⋅ 3 3 − 2 3 ⋅ 3 2 ) : ( − 36 ) (2x−15) 3 =(2 2 ⋅3 3 −2 3 ⋅3 2 ):(−36) => ( 2 x − 15 ) 3 = ( 4 ⋅ 27 − 8 ⋅ 9 ) : ( − 36 ) (2x−15) 3 =(4⋅27−8⋅9):(−36) => ( 2 x − 15 ) 3 = ( 108 − 72 ) − 36 = 36 − 36 = − 1 (2x−15) 3 = −36 (108−72) = −36 36 =−1 =>2x-15=-1 =>2x=14 =>x=7 c: ( 4 x + 1 ) 4 = ( 4 x + 1 ) 6 (4x+1) 4 =(4x+1) 6 => ( 4 x + 1 ) 6 − ( 4 x + 1 ) 4 = 0 (4x+1) 6 −(4x+1) 4 =0 => ( 4 x + 1 ) 4 ⋅ [ ( 4 x + 1 ) 2 − 1 ] = 0 (4x+1) 4 ⋅[(4x+1) 2 −1]=0 => ( 4 x + 1 ) 4 ( 4 x + 1 − 1 ) ( 4 x + 1 + 1 ) = 0 (4x+1) 4 (4x+1−1)(4x+1+1)=0 => 4 x ( 4 x + 2 ) ( 4 x + 1 ) 4 = 0 4x(4x+2)(4x+1) 4 =0 => [ x = 0 4 x + 2 = 0 4 x + 1 = 0 ⇔ [ x = 0 x = − 1 2 x = − 1 4 x=0 4x+2=0 4x+1=0 ⇔ x=0 x=− 2 1 x=− 4 1 d: ( 1 1 ⋅ 2 + 1 2 ⋅ 3 + . . . + 1 99 ⋅ 100 ) ⋅ x 2 = 99 ( 1⋅2 1 + 2⋅3 1 +...+ 99⋅100 1 )⋅x 2 =99 => ( 1 − 1 2 + 1 2 − 1 3 + . . . + 1 99 − 1 100 ) ⋅ x 2 = 99 (1− 2 1 + 2 1 − 3 1 +...+ 99 1 − 100 1 )⋅x 2 =99 => ( 1 − 1 100 ) ⋅ x 2 = 99 (1− 100 1 )⋅x 2 =99 => x 2 = 99 : 99 100 = 100 x 2 =99: 100 99 =100 => x = ± 10 x=±10 2: 2xy-x-y=2 => x ( 2 y − 1 ) − y + 1 2 = 2 + 1 2 x(2y−1)−y+ 2 1 =2+ 2 1 => 2 x ( y − 1 2 ) − ( y − 1 2 ) = 5 2 2x(y− 2 1 )−(y− 2 1 )= 2 5 => ( 2 x − 1 ) ( y − 1 2 ) = 5 2 (2x−1)(y− 2 1 )= 2 5 =>(2x-1)(2y-1)=5 => ( 2 x − 1 ; 2 y − 1 ) ∈ { ( 1 ; 5 ) ; ( 5 ; 1 ) ; ( − 1 ; − 5 ) ; ( − 5 ; − 1 ) } (2x−1;2y−1)∈{(1;5);(5;1);(−1;−5);(−5;−1)} => ( x ; y ) ∈ { ( 1 ; 3 ) ; ( 3 ; 1 ) ; ( 0 ; − 2 ) ; ( − 2 ; 0 ) } (x;y)∈{(1;3);(3;1);(0;−2);(−2;0)}

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