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a, \(M=\dfrac{1}{2}x^4y^4\)
b, hệ số : 1/2 ; biến x^4y^4 ; bậc 8
Giải:
a) \(A=\left(-\dfrac{2}{17}x^3y^5\right).\dfrac{34}{5}x^2y\)
\(\Leftrightarrow A=-\dfrac{2}{17}x^3y^5.\dfrac{34}{5}x^2y\)
\(\Leftrightarrow A=\left(-\dfrac{2}{17}.\dfrac{34}{5}\right)\left(x^3.x^2\right)\left(y^5.y\right)\)
\(\Leftrightarrow A=-\dfrac{4}{5}x^5y^6\)
Bậc của đơn thức A: \(5+6=12\)
b) Tại x = -1 và y = -1, giá trị của A là:
\(\Leftrightarrow A=-\dfrac{4}{5}\left(-1\right)^5\left(-1\right)^6\)
\(\Leftrightarrow A=-\dfrac{4}{5}\left(-1\right)^{11}\)
\(\Leftrightarrow A=-\dfrac{4}{5}.\left(-1\right)\)
\(\Leftrightarrow A=\dfrac{4}{5}\)
Vậy ...
a: \(A=\dfrac{19}{5}xy^2\cdot x^3y=\dfrac{19}{5}x^4y^3\)
b: Hệ số là 19/5
Bậc là 7
c: Khi x=1 và y=2 thì \(A=\dfrac{19}{5}\cdot1^4\cdot2^3=\dfrac{152}{5}\)
\(M=\left(\frac{-4}{3}x^2y\right)\left(\frac{15}{2}xy^3\right)\left(2020x^2y^3\right)^0\)
\(M=\left(\frac{-4}{3}.\frac{15}{2}\right)\left(x^2.x\right)\left(y.y^3\right).1\)
\(M=-10x^3y^4\)
1) a)
=\(\left(4-1+8\right)x^2=11x^2\)
b) =\(\left(\dfrac{1}{2}-\dfrac{3}{4}+1\right)x^2y^2=\dfrac{3}{4}x^2y^2\)
c) =(3-7+4-6)y=5y 2) a) ...=\(\left[\left(\dfrac{-2}{3}y^3\right)-\dfrac{1}{2}y^3\right]+3y^2-y^2\\ =\left[\left(\dfrac{-2}{3}-\dfrac{1}{2}\right)y^3\right]+\left(3-1\right)y^2=\dfrac{-7}{6}y^3+2y^2\) b) ...=\(\left(5x^3-x^3\right)-\left(3x^2+4x^2\right)+\left(x-x\right)=4x^3-7x^2\) 3) a)A=\(\left(5.\dfrac{1}{2}\right).\left(x.x^2.x\right)\left(y^2.y^2\right)=\dfrac{5}{2}x^4y^4\) b)Vậy Đơn thức A có bậc 8; hệ số là \(\dfrac{5}{2}\); phần biến là \(x^4y^4\) c)Khi x=1;y=-1 thì A=\(\dfrac{5}{2}.1^4.\left(-1\right)^4=\dfrac{5}{2}\)
1.\(A=-\dfrac{3}{4}x^2yz;B=\dfrac{1}{3}xy^2;C=-\dfrac{8}{7}xy^2\)
\(A.\left(B+C\right)=-\dfrac{3}{4}x^2yz\left[\dfrac{1}{3}xy^2+\left(-\dfrac{8}{7}xy^2\right)\right]\)
\(=-\dfrac{3}{4}x^2yz\left(\dfrac{1}{3}xy^2-\dfrac{8}{7}xy^2\right)\)
\(=\left(-\dfrac{3}{4}x^2yz\right)\dfrac{1}{3}xy^2-\left(-\dfrac{3}{4}x^2yz\right)\dfrac{8}{7}xy^2\)
\(=-\dfrac{1}{4}x^3y^3z+\dfrac{6}{7}x^3y^3z\)
1. Ta có: \(-\dfrac{3}{4}x^2yz;B=\dfrac{1}{3}xy^2;C=-\dfrac{8}{7}xy^2\)
\(B+C=\dfrac{1}{3}xy^2-\dfrac{8}{7}xy^2=-\dfrac{17}{21}xy^2\)
\(A.\left(B+C\right)=\left(-\dfrac{3}{4}x^2yz\right).\left(-\dfrac{17}{21}xy^2\right)\)
\(\Rightarrow A.\left(B+C\right)=\dfrac{17}{28}x^3y^3z\)
a) Ta có: \(A=1\dfrac{1}{4}\cdot x^3y\cdot\left(-\dfrac{6}{7}xy^5\right)^0\cdot\left(-2\dfrac{2}{3}xy\right)\)
\(=\dfrac{5}{4}x^3y\cdot\dfrac{-8}{3}xy\)
\(=\left(\dfrac{5}{4}\cdot\dfrac{-8}{3}\right)\cdot\left(x^3\cdot x\right)\cdot\left(y\cdot y\right)\)
\(=\dfrac{-10}{3}x^4y^2\)
\(M=\dfrac{49}{9}x^6y^2\)
\(M=(-\dfrac{7}{3})^2(x^3y)^2(-2020x^{15}y^{17}2^{19}t^{1000})^0\)
\(M=\dfrac{49}{9}x^2y^2.x^6y^2.1\)
\(M=\dfrac{49}{9}(x^2x^6).(y^2y^2)\)
\(M=\dfrac{49}{9}x^8y^4\)