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39 광고문의 0422 258 092, 0432 008 985 visionweekly01@gmail.com Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answwer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy  xy = 200 ① perimeter of the rectangle = 2 ( width + length ) 60 = 2 ( x + y )  x + y = 30 ② From ① xy = 200 = 1 × 200 = 2 × 100 = 4 × 50 = 8 × 25 = 10 × 20 If x = 10 and y = 20 , the equation of ② is satisfied. The dimensions of the rectangle is 10 × 20. ( width = 10 and length = 20 ) 2022 역사와전통을자랑하는호주최고의제임스안학원이 미래를준비하며열심히노력하는학생여러분을기다립니다. 김선생 연습문제&풀이 Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answeerr ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 0 = 3 2 = 4 ∴ = 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 ( 2 − 9 − 10) 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy  xy = 200 ① perimeter of the rectangle = 2 ( width + length ) 60 = 2 ( x + y )  x + y = 30 ② From ① xy = 200 = 1 × 200 = 2 × 100 = 4 × 50 = 8 × 25 = 10 × 20 If x = 10 and y = 20 , the equation of ② is satisfied. The dimensions of the rectangle is 10 × 20. ( width = 10 and length = 20 ) Solve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The a ea of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 3 (2 − ) = 2 ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 = (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) = 2 + 2 − 4 × 2( − 2 + 1) = 4 + 2 − 4 ( − 2 + 1) (9) Let x = width of the rectangle y = length of the rectangle area of the rectangle = width × length 200 = xy  xy = 200 ① perimeter of the rectangle = 2 ( width + length ) 60 = 2 ( x + y )  x + y = 30 ② From ① xy = 200 = 1 × 200 = 2 × 100 = 4 × 50 = 8 × 25 = 10 × 20 Solvve the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 ) Factorrise the followings (5-8) 5 5 10 20 ( 6, 7 학년 ) 6 − + 6 − 3 + ( 8 학년 ) 7 2 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 5 + 40 − 16 2 ( 10 학년 ) 9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 3 = 8 − 4 = 8 ∴ − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + = − 36 6 = − 36 ∴ = − 6 3 3 − − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) 10 + 1 8 4 2 − 12 + 9 − 25 + 40 − 16 2 = (2 ) 2 − 2 2 3 + 3 2 − (25 − 40 + 16 2 ) = (2 − 3) 2 − 5 2 − 2 5 (4 ) + (4 ) 2 (2 − 3) 2 − (5 − 4 ) 2 = (2 − 3) − 5 − 4 {(2 − 3) + 5 − 4 } = 2 − 3 − 5 + 4 (2 − 3 + 5 − 4 ) = 2 − 8 + 4 (2 + 2 − 4 ) 2 + 2 − 4 × 2( − + 1) 4 + 2 4 ( − 2 + 1) (9) Let x = width of the rectangle y length of the rectangle Solve the foollowings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followinngs (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 = 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 = − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) = 2 − ( + 3 ) 7 2 − 9 − 10 = ( 2 − 9 − 10) = − 10 + 1 Solvee the followings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followings (5-8) 5 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 − 9 − 10 ( 9 학년 ) 8 4 2 12 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = 4 − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 4 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = 2 , 1/3 5 5 − 10 − 20 5 × − 5 × 2 − 5 × 4 = 5 ( − 2 − 4 ) 6 − + 6 − 3 + 2 − + 2 + 6 − 3 = − + 2 + 3 (2 − ) = 2 − + 3 (2 − ) Solve the followings (1-4) 1 2 − 3 − (5 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 −2 + 4 − −2 + = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 − 4 − 2 = 0 ( 8, 9 학년 ) 4 3 − 7 + 2 = 0 ( 9, 10 학년 ) Factorise the followwings (5-8) 5 5 − 10 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 0 ( 9 학년 ) 8 4 2 − 2 + 9 − 5 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 200 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 − 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 − 3 = 8 − 4 = 8 ∴ = − 2 3 − 2 + 4 − − 2 + = 4 − 0 − 3 + 4 3 − − = 4 − 40 − 3 + 4 6 + = 36 6 + = − 36 6 = 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 4 − 2 = 0 = 3 2 = 4 ∴ = 2 , 3 4 3 2 − 7 + 2 = 0 − 2 3 − 1 = 0 − 2 = 0 3 − 1 = 0 = 2 3 = 1 ∴ = , 1/3 5 5 10 − 20 Solve the followiings (1-4) 1 − 2 − 3 − (5 − 3) = 3 + 8 ( 6 , 7 학년 ) 2 3 2 + 4 − −2 = 4 − 10 − 3 + 4 ( 7, 8 학년 ) 3 3 4 − 2 = 0 ( 8, 9 학년 ) 4 3 2 − 7 + 2 = 0 ( 9, 10 학년 ) Factoorise the followings (5-8) 5 5 1 − 20 ( 6, 7 학년 ) 6 − + 6 − 3 + 2 ( 8 학년 ) 7 2 9 − 0 ( 9 학년 ) 8 4 2 − 12 + 9 − 25 + 40 − 16 2 ( 10 학년 ) (9) The area of a rectangle is 00 2 and its perimeter is 60 m. Find its dimensions of the rectangle. 김선생수학알제브라연습문제 (268) ( Exercise of Algebra ) ( Answer ) 1 2 − 3 − (5 − 3) = 3 + 8 − ( − 1) − 2 = 3 + 8 − 2 3 = 8 − 4 = 8 ∴ = − 2 2 3 − 2 + 4 − − 2 + = − 10 − 3 + 4 3 2 − − = 4 − 40 − 3 + 6 + = − 36 6 + − = − 36 6 = − 36 ∴ = − 6 3 3 − 4 − 2 = 0 3 − = 0 2 = 0 = 3 2 = 4 ∴ = 2 , 3 ⓒ본광고이미지는비전매거진이제작하였습니다. *정답은40p에있습니다.