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a, \(2x^2y^2.\frac{1}{4}xy^3.\left(-3xy\right)\)
=\(\left[2.\frac{1}{4}.\left(-3\right)\right].\left(x^2.x.x\right)\left(y^2.y^3.y\right)\)
= \(\left(\frac{-3}{2}\right).x^4.y^6\)
= \(\frac{-3}{2}x^4y^6\)
\(\frac{-3}{2}\) là hệ số, \(x^4y^6\) là biến
b, \(\left(-2x^3y\right)^2.xy^2.\frac{1}{5}y^5\)
= \(\left[\left(-2\right)^2.1.\frac{1}{5}\right].\left(x^6.x\right).\left(y^2.y^2.y^5\right)\)
= \(\frac{4}{5}.x^7.y^9\)
= \(\frac{4}{5}x^7y^9\)
\(\frac{4}{5}\) là hệ số, \(x^7y^9\) là biến
Không chắc sẽ làm đung toàn bộ nhé '-'
a)\(2x^2y^2.\frac{1}{4}xy^3\left(-3xy\right)\)
= \(\left[2.\frac{1}{4}.\left(-3\right)\right]\).\(\left(x^2.x.x\right).\left(y^2.y^3.y\right)\)
=\(\frac{-2}{3}x^4y^6\)
Phần hệ số:\(\frac{-2}{3}\)
Phần biến:10
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 :
\(\frac{x}{18}=\frac{y}{9}\)\(\Leftrightarrow\)\(\frac{x}{2}=y\)
\(\Rightarrow\)\(x=2y\)
Thay \(x=2y\) vào \(A=\frac{2x-3y}{2x+3y}\) ta được :
\(A=\frac{2.2y-3y}{2.2y+3y}=\frac{4y-3y}{4y+3y}=\frac{y}{7y}=\frac{1}{7}\)
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\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
a/ \(x^2+y^2=0\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) \(\Rightarrow A=0\)
b/ Do \(x=19\Rightarrow20=x+1\)
\(B=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+20\)
\(B=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+20\)
\(B=20-x=20-19=1\)
c/ \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
\(C=\frac{\left(x+y\right)}{y}.\frac{\left(y+z\right)}{z}.\frac{\left(x+z\right)}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=\frac{-xyz}{xyz}=-1\)