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\(A=x^2y^3\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)=\dfrac{67}{60}x^2y^3\)
\(B=x^6y^3\cdot\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
\(A+B=\dfrac{67}{60}x^2y^3+\dfrac{1}{4}x^8y^7z^2\)
\(A-B=\dfrac{67}{60}x^2y^3-\dfrac{1}{4}x^8y^7z^2\)
A=x2y3(15+23−34+1)=6760x2y3A=x2y3(15+23−34+1)=6760x2y3
B=x6y3⋅14x2y4z2=14x8y7z2B=x6y3⋅14x2y4z2=14x8y7z2
A+B=6760x2y3+14x8y7z2A+B=6760x2y3+14x8y7z2
A−B=6760x2y3−14x8y7z2
\(A=\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3=\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)x^2y^3=\dfrac{67}{60}x^2y^3\\ B=\left(x^2y\right)^3\left(\dfrac{1}{2}xy^2z\right)^2=x^6y^3.\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
2: Thay \(x=\dfrac{1}{2}\) và y=2 vào M, ta được:
\(M=\dfrac{2\cdot\left(\dfrac{1}{2}\right)^2\cdot2-1.2\cdot\left(3\cdot\dfrac{1}{2}-2\cdot2\right)}{\dfrac{1}{2}\cdot2}\)
\(=4\cdot\dfrac{1}{4}-1.2\left(\dfrac{3}{2}-4\right)\)
\(=1-1.8+4.8\)
\(=4\)
1: Ta có: \(\left(-\dfrac{2}{3}x^3y^2\right)z\cdot5xy^2z^2\)
\(=\left(-\dfrac{2}{3}\cdot5\right)\cdot\left(x^3\cdot x\right)\cdot\left(y^2\cdot y^2\right)\cdot\left(z\cdot z^2\right)\)
\(=\dfrac{-10}{3}x^4y^4z^3\)
7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36
Nên theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6
\(\Rightarrow\)x=6.5=30
y=6.6=36
z=6.7=42
vậy x=30,y=36,z=42
\(\frac{1}{2}x^2y^3-x^2y^3+3x^2y^2z^2-z^4-3x^2y^2z^2\)
\(=\left(\frac{1}{2}x^2y^3-x^2y^3\right)+\left(3x^2y^2z^2-3x^2y^2z^2\right)-z^4\)
\(=-\frac{1}{2}x^2y^3-z^4\)
a) x2+5x2+(−3x2)=3x2
b) 5xy2+12xy2+14xy2+(−12)xy2=19xy2
c) 3x2y2z2+x2y2z2=4x2y2z2
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=\dfrac{3x-2y+2z}{2\cdot3-3\cdot2+2\cdot6}=\dfrac{24}{12}=2\\ \Rightarrow\left\{{}\begin{matrix}x=24\\y=36\\z=72\end{matrix}\right.\)
a) x6+x2y5+xy6+x2y5-xy6
= x6+(x2y5+x2y5)+(xy6-xy6)
= x6+2x2y5
b) \(\dfrac{1}{2}\)x2y3-x2y3+3x2y2z2-z4-3x2y2z2
= (\(\dfrac{1}{2}\)x2y3-x2y3)+(3x2y2z2-3x2y2z2)-z4
= -\(\dfrac{1}{2}\)x2y3-z4