Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\dfrac{2020}{2021}xy^5z\right).\left(\dfrac{2020}{2021}x^3yz^2\right).\left(-\dfrac{2020}{2021}\right)^0\)
\(a)A=\dfrac{2020.2021.2020}{2021.2020.2021}.\left(x.x^3\right).\left(y^5.y\right).\left(z.z^2\right)\Leftrightarrow A=\dfrac{2020}{2021}x^4.y^6.z^3\)
\(b)A=\dfrac{2020}{2021}x^4.y^6.z^3\)
\(\Rightarrow\text{A có hệ số là:}\dfrac{2020}{2021}\)
\(\text{Phần biến là:}\left(x,y,z\right)\)
\(c)\text{Xét A ta có:}\dfrac{2020}{2021}< 0;x^4,y^6\text{ luôn }< 0\)
\(\Rightarrow\dfrac{2020}{2021}x^4.y^6>0\Rightarrow\text{ Nếu }z< 0\Rightarrow A\le0\text{ và z có số mũ là:3}\)
\(\text{Chẳng hạn:}\left(-\right).\left(-\right).\left(-\right)=\left(-\right).< 0\Rightarrow z\text{ phải }\ge0\text{ thì }A\ge0\)
\(\Rightarrow Z\in N\)
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
Bài làm :
\(\frac{8}{9}-\left(\frac{-1}{3}\right)^2+\left(\frac{5}{6}\right)^{2020}\times\left(\frac{6}{5}\right)^{2020}\)
\(=\frac{8}{9}-\frac{1}{9}+\left(\frac{5}{6}\times\frac{6}{5}\right)^{2020}\)
\(=\frac{7}{9}+1\)
\(=\frac{16}{9}\)
Học tốt nhé
Khẳng định a là khẳng định đúng