Giải bài 7.14 trang 33 SGK Toán 7 tập 2 – Kết nối tri thức

Đề bài

Cho hai đa thức:

\(A = 6{x^4} – 4{x^3} + x – \dfrac{1}{3};B =  – 3{x^4} – 2{x^3} – 5{x^2} + x + \dfrac{2}{3}\)

Tính A + B và A – B

Lời giải chi tiết

Cách 1:

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

Cách 2: