3. Gọi đơn thức phải tìm là \(B=ax^p.y^q\), ta có:

\(5.x^{2+n}.y^{m-3}=\left(-\dfrac{3}{2}x^ny^3\right).ax^py^q\)

\(\Rightarrow5x^{2+n}y^{m-3}=-\dfrac{3}{2}ax^{n+p}y^{q+3}\)

Từ đây suy ra:

\(5=-\dfrac{3}{2}a\Rightarrow a=-\dfrac{10}{3}\)

\(n+p=n+2\Rightarrow p=2\)

\(q+3=m-3\Rightarrow q=m-6\)

Vậy đơn thức phải tìm là:

\(B=-\dfrac{10}{3}x^2y^{m-6}\)

2. a) Ta có: \(3x^3y^2=\dfrac{3}{5}x^3y^2+a.x^3y^2\)

\(\Rightarrow3=\dfrac{3}{5}+a\)

\(\Rightarrow a=3-\dfrac{3}{5}=\dfrac{12}{5}\)

Vậy: \(3x^3y^2=\dfrac{3}{5}x^3y^2+\dfrac{12}{5}x^3y^2\)

b) \(3x^3y^2=bx^3y^2-\left(-\dfrac{1}{4}x^3y^2\right)\)

\(\Rightarrow3=b+\dfrac{1}{4}\)

\(\Rightarrow b=3-\dfrac{1}{4}=\dfrac{11}{4}\)

Vậy: \(3x^3y^2=\dfrac{11}{4}x^3y^2-\left(-\dfrac{1}{4}x^3y^2\right)\)

c) Ta có: \(3x^3y^2=\left(-\dfrac{5}{7}x^2y\right).F\)

\(\Rightarrow F=\left(3x^3y^2\right):\left(-\dfrac{5}{7}x^2y\right)\)

\(\Rightarrow F=\left(3:\left(-\dfrac{5}{7}\right)\right)\dfrac{x^3y^2}{x^2y}\)

\(\Rightarrow F=-\dfrac{21}{5}xy\)

Vậy: \(3x^3y^2=\left(-\dfrac{5}{7}x^2y\right).\left(\dfrac{-21}{5}xy\right)\)

1. Ta có: \(-\dfrac{3}{4}x^2yz;B=\dfrac{1}{3}xy^2;C=-\dfrac{8}{7}xy^2\)

\(B+C=\dfrac{1}{3}xy^2-\dfrac{8}{7}xy^2=-\dfrac{17}{21}xy^2\)

\(A.\left(B+C\right)=\left(-\dfrac{3}{4}x^2yz\right).\left(-\dfrac{17}{21}xy^2\right)\)

\(\Rightarrow A.\left(B+C\right)=\dfrac{17}{28}x^3y^3z\)

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a) Ta có: \(\left(\dfrac{1}{3}\right)^{2n-1}=243\Rightarrow\left(3^{-1}\right)^{2n-1}=3^5\)

\(\Rightarrow3^{-2n+1}=3^5\)

Suy ra: \(-2n+1=5\Rightarrow-2n=4\Rightarrow n=-2\)

b) \(\left(0,125\right)^{n+1}=64\Rightarrow\left(\dfrac{1}{8}\right)^{n+1}=8^2\)

\(\Rightarrow\left(8^{-1}\right)^{n+1}=8^2\)

\(\Rightarrow8^{-n-1}=8^2\)

Suy ra: \(-n-1=2\Rightarrow n=-3\)

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