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Solving Exercises Lesson 26 Addition and subtraction of one-variable polynomials (Chapter 7 Math 7 Connect)
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Solve problem 7.12, page 33, Math 7 textbook Connecting knowledge volume 2 – KNTT

Detailed instructions for solving Problem 7.12

Solution method

Remove the brackets and group the terms of the same degree.

Detailed explanation

We have: (x² – 3x + 2) + (4x .)³ – x² + x – 1)

= x² – 3x + 2 + 4x³ – x² + x – 1

= 4x³ + (x² – x² ) + (-3x + x) + (2 – 1)

= 4x³ – 2x + 1

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Solve problem 7.13 page 33 Math 7 Textbook Connecting knowledge volume 2 – KNTT

Detailed instructions for solving problem 7.13

Solution method

Set the subtraction so that the terms of the same degree are placed in the same column, and then subtract each column.

Detailed explanation

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Solve problem 7.14, page 33, Math 7 Textbook Connecting knowledge volume 2 – KNTT

Detailed instructions for solving Problem 7.14

Solution method

Method 1: Remove the brackets and group the terms of the same degree.

Method 2: Set the addition (subtract) so that the terms of the same degree are placed in the same column and then add (subtract ) by each column.

Detailed explanation

Method 1:

\(\begin{array}{l}A + B = (6{x^4} – 4{x^3} + x – \dfrac{1}{3}) + ( – 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} – 4{x^3} + x – \dfrac{1}{ 3} – 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3}\\ = (6{x^4} – 3{x^4 }) + ( – 4{x^3} – 2{x^3}) – 5{x^2} + (x + x) + ( – \dfrac{1}{3} + \dfrac{2}{ 3})\\ = 3{x^4} – 6{x^3} – 5{x^2} + 2x + \dfrac{1}{3}\\A – B = (6{x^4} – 4{x^3} + x – \dfrac{1}{3}) – ( ​​– 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2} {3})\\ = 6{x^4} – 4{x^3} + x – \dfrac{1}{3} + 3{x^4} + 2{x^3} + 5{x^ 2} – x – \dfrac{2}{3}\\ = (6{x^4} + 3{x^4}) + ( – 4{x^3} + 2{x^3}) + 5 {x^2} + (x – x) + ( – \dfrac{1}{3} – \dfrac{2}{3})\\ = 9{x^4} – 2{x^3} + 5 {x^2} – 1\end{array}\)\(\begin{array}{l}A + B = (6{x^4} – 4{x^3} + x – \dfrac{1} {3}) + ( – 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} – 4 {x^3} + x – \dfrac{1}{3} – 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3}\\ = (6{x^4} – 3{x^4}) + ( – 4{x^3} – 2{x^3}) – 5{x^2} + (x + x) + ( – \ dfrac{1}{3} + \dfrac{2}{3})\\ = 3{x^4} – 6{x^3} – 5{x^2} + 2x + \dfrac{1}{3}\\A – B = (6{x^4} – 4{x^3} + x – \dfrac{1}{3}) – ( ​​– 3{x^ 4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3})\\ = 6{x^4} – 4{x^3} + x – \dfrac{ 1}{3} + 3{x^4} + 2{x^3} + 5{x^2} – x – \dfrac{2}{3}\\ = (6{x^4} + 3{ x^4}) + ( – 4{x^3} + 2{x^3}) + 5{x^2} + (x – x) + ( – \dfrac{1}{3} – \dfrac{ 2}{3})\\ = 9{x^4} – 2{x^3} + 5{x^2} – 1\end{array}\)

Method 2:

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Solve problem 7.15 page 33 Math textbook 7 Connecting knowledge volume 2 – KNTT

Detailed instructions for solving problem 7.15

Solution method

Remove the brackets and group the terms of the same degree.

Detailed explanation

\(\begin{array}{l}A + B + C = (3{x^4} – 2{x^3} – x + 1) + ( – 2{x^3} + 4{x^2 } + 5x) + ( – 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} – 2{x^3} – x + 1 – 2{x^3} + 4{x^2} + 5x – 3{x^4} + 2{x^2} + 5\\ = (3{x^4} – 3{x^4}) + ( – 2{x^ 3} – 2{x^3}) + (4{x^2} + 2{x^2}) + ( – x + 5x) + (1 + 5)\\ = 0 + ( – 4{x^) 3}) + 6{x^2} + 4x + 6\\ = – 4{x^3} + 6{x^2} + 4x + 6\\A – B + C = (3{x^4} – 2{x^3} – x + 1) – ( ​​– 2{x^3} + 4{x^2} + 5x) + ( – 3{x^4} + 2{x^2} + 5) \\ = 3{x^4} – 2{x^3} – x + 1 + 2{x^3} – 4{x^2} – 5x – 3{x^4} + 2{x^2} + 5\\ = (3{x^4} – 3{x^4}) + ( – 2{x^3} + 2{x^3}) + ( – 4{x^2} + 2{x ^2}) + ( – x – 5x) + (1 + 5)\\ = 0 + 0 + ( – 2{x^2}) – 6x + 6\\ = – 2{x^2} – 6x + 6\\A – B – C = (3{x^4} – 2{x^3} – x + 1) – ( ​​– 2{x^3} + 4{x^2} + 5x) – ( ​​– 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} – 2{x^3} – x + 1 + 2{x^3} – 4{x^2} – 5x + 3{x^4} – 2{x^2} – 5\\ = (3{x^4} + 3{x^4}) + ( – 2{x^3} + 2{x^ 3}) + ( – 4{x^2} – 2{x^2}) + ( – x – 5x) + (1 – 5)\\ = 6{x^4} + 0 + ( – 6{x ^2}) – 6x + ( – 4)\\ = 6{x^4} – 6{x^2} – 6x – 4\end{array}\)

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Solve problem 7.16 page 33 Math 7 textbook Connecting knowledge volume 2 – KNTT

Detailed instructions for solving Problem 7.16

Solution method

Write a polynomial representing the amount

Money to buy 1 type of book = number of books. price of a book

Detailed explanation

a) The polynomial representing the amount Nam has to pay for the comic is: A = (x +5). 15 000 = 15 000x + 75 000 (VND)

The polynomial representing the amount Nam has to pay for the reference book is: B = (x + 8) . 12 500 = 12 500x + 100 000 (VND)

The polynomial representing the amount Nam has to pay for science books is: C = x . 21 500 (VND)

b) The polynomial expressing the total amount Nam has to pay to buy that number of books is:

P = A + B + C = = 15 000x + 75 000 + 12 500x + 100 000 + x . 21 500

= (15 000 + 12 500 + 21 500)x + (75 000 + 100 000)

= 49 000x + 175 000 (VND)

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Solve problem 7.17 page 33 Math textbook 7 Connecting knowledge volume 2 – KNTT

Detailed instructions for solving Problem 7.17

Solution method

+ Indicates the length and width of the rectangles

Area of ​​rectangle = length. width

Detailed explanation

a) The pool has a length of 3x and a width of x, so the polynomial for the area of ​​the pool is:

B = 3x. x = 3.x²

b) The plot of land has a length of 65 and a width of 5 + x + 4 = x + 9, so the polynomial representing the area of ​​​​the land is:

D = 65. (x+9) = 65x + 585

c) The area around the swimming pool = the area of ​​​​the land – the area of ​​​​the swimming pool, so the polynomial expressing the area of ​​​​the land around the swimming pool is:

Q = D – B = 65x + 585 – 3.x² = -3.x² +65x + 585

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