Làm tính chia :
a) \(\left(\dfrac{5}{7}x^2y\right)^3:\left(\dfrac{1}{7}xy\right)^3\)
b) \(\left(-x^3y^2z\right)^4:\left(-xy^2z\right)^3\)
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a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)
c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)
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: \(\left(-\dfrac{2}{3}x^3y^2\right)z\cdot5xy^2z^2\)
\(=\left(-\dfrac{2}{3}\cdot5\right)\cdot\left(x^3\cdot x\right)\cdot\left(y^2\cdot y^2\right)\cdot\left(z\cdot z^2\right)\)
\(=\dfrac{-10}{3}x^4y^4z^3\)
\(A=x^2y^3\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)=\dfrac{67}{60}x^2y^3\)
\(B=x^6y^3\cdot\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
\(A+B=\dfrac{67}{60}x^2y^3+\dfrac{1}{4}x^8y^7z^2\)
\(A-B=\dfrac{67}{60}x^2y^3-\dfrac{1}{4}x^8y^7z^2\)
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
\(A=\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3=\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)x^2y^3=\dfrac{67}{60}x^2y^3\\ B=\left(x^2y\right)^3\left(\dfrac{1}{2}xy^2z\right)^2=x^6y^3.\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
a, \(\left(-2xy^2z^3\right)^3.\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)
\(=\left(-2\right)^3.x^3.\left(y^2\right)^3.z^3.\left(\dfrac{5}{2}\right)^2.x^2.\left(y^3\right)^2.\dfrac{-4}{125}.x.y\)
\(=\left(-2\right)^3.\left(\dfrac{5}{2}\right)^2.\dfrac{-4}{125}.\left(x^3.x^2.x\right).\left(y^6.y^6.y\right).z^3\)
\(=\left(-8\right).\dfrac{25}{4}.\dfrac{-4}{125}.x^6.y^{13}.z^3\)
\(=1,6.x^6.y^{13}.z^3\)
a, \(\left(-2xy^2z^3\right).\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)
= \(\left(-5x^2y^5z^3\right)^5.\left(\dfrac{-4}{125}xy\right)\)
= \(\left(\dfrac{4}{25}x^3y^6z^3\right)^5\)
b, \(2\dfrac{1}{3}x^2y^5-3\dfrac{2}{5}x^3y-1\dfrac{1}{2}x^2y^5+2\dfrac{2}{3}x^3y\)
= \(\dfrac{7}{3}x^2y^5-\dfrac{17}{5}x^3y-\dfrac{3}{2}x^2y^5+\dfrac{8}{3}x^3y\)
= \(\dfrac{5}{6}x^2y^5-\dfrac{11}{15}x^3y\)
a)\(\left(\dfrac{5}{7}x^2y\right)^3:\left(\dfrac{1}{7}xy\right)^3=\dfrac{125}{343}x^5y^3:\dfrac{1}{343}x^3y^3=\left(\dfrac{125}{343}:\dfrac{1}{343}\right)\left(x^5:x^3\right)\left(y^3:y^3\right)=125x^2\)
b)\(\left(-x^3y^2z\right)^4:\left(-xy^2z\right)^3=x^7y^6z^4:x^3y^5z^3=\left(x^7:x^3\right)\left(y^6:y^5\right)\left(z^4:z^3\right)=x^4yz\)