Question

[Minh Nhật]

Giúp mik giải toán nhé 

Answer 1

[✧ ( •̀ ω •́ )✧₸ĦV✧( •̀ ω •́ )✧]

\(\text{Bài 1)}\)

\(A=x^3\left(-\frac{5}{4}x^2y\right).\left(\frac{2}{5}x^3y^4\right)\)

\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(-\frac{5}{4}.\frac{2}{5}\right)\)

\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)

\(B=\left(-\frac{3}{4}x^5y^4\right).\left(xy^2\right).\left(-\frac{8}{9}x^2y^5\right)\)

\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y^5\right).\left(-\frac{3}{4}.-\frac{8}{9}\right)\)

\(B=x^8.y^{11}.\frac{2}{3}\)

\(\text{Bài 2)}\)

\(a)\text{Bậc của đa thức là:7}\)

\(b)\text{Thay x=-1;y=-2 vào biểu thức A,ta được:}\)

\(A=2.\left(-1\right)^2.\left(-2\right)^3-4,5.\left(-1\right)^3.\left(-2\right)^2+3.\left(-1\right)^4.\left(-2\right)^3-6.\left(-1\right).\left(-2\right)^3\)

\(A=2.1.\left(-8\right)-4,5.\left(-1\right).4+3.1.\left(-8\right)-6.\left(-1\right).\left(-8\right)\)

\(A=\left(-16\right)-\left(-18\right)+\left(-24\right)-48\)

\(A=2+\left(-24\right)-48\)

\(A=\left(-22\right)-48=-70\)

\(\text{Vậy giá trị của biểu thức A tại x=-1;y=-2 là:-70}\)

\(\text{Bài 3)}\)

\(A+B=\left(4x^2-5xy+3y^2\right)+\left(3x^2+2xy-y^2\right)\)

             \(=4x^2-5xy+3y^2+3x^2+2xy-y^2\)

             \(=\left(4x^2+3x^2\right)+\left(5xy+2xy\right)+\left(3y^2-y^2\right)\)

            \(=7x^2+7xy+2y^2\)

\(A-B=\left(4x^2-5xy+3y^2\right)-\left(3x^2+2xy-y^2\right)\)

            \(=4x^2-5xy+3y^2-3x^2-2xy+y^2\)

            \(=\left(4x^2-3x^2\right)+\left(5xy-2xy\right)+\left(3y^2+y^2\right)\)

            \(=x^2+3xy+4y^2\)

\(\text{Bài 4)}\)

\(a)M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(\Rightarrow M=\left(6x^2+9xy-y^2\right)-\left(5x^2-2xy\right)\)

\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)

\(\Rightarrow M=\left(6x^2-5x^2\right)+\left(9xy+2xy\right)+y^2\)

\(\Rightarrow M=x^2+11xy+y^2\)