-3x/2020x+4040/=15
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ST
1
TM
12 tháng 4 2021
\(M=\left(\frac{-4}{3}x^2y\right)\left(\frac{15}{2}xy^3\right)\left(2020x^2y^3\right)^0\)
\(M=\left(\frac{-4}{3}.\frac{15}{2}\right)\left(x^2.x\right)\left(y.y^3\right).1\)
\(M=-10x^3y^4\)
7 tháng 9 2021
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
16 tháng 9 2021
\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)
\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)
VN
1
21 tháng 6 2021
f(x) = \(\left(x^6-2019x^5\right)-\left(x^5-2019x^4\right)+\left(x^4-2019x^3\right)-\left(x^3-2019x^2\right)+\left(x^2-2019x\right)-\left(x-2019\right)+1\)
= \(x^5\left(x-2019\right)-x^4\left(x-2019\right)+x^3\left(x-2019\right)-x^2\left(x-2019\right)+x\left(x-2019\right)-\left(x-2019\right)+1\)
Thay x = 2019 vào f(x), ta có:
f(2019) = 0 + 0 + 0 + 0 + 0 +0 + 1 = 1
NH
1
8 tháng 7 2021
\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)
\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)
\(=-x+2000=-2019+2000\)
\(=-19\)
KN
8 tháng 11 2019
\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)
\(=x^{2019}-2019x^{2018}-x^{2018}+2019x^{2017}+x^{2017}\)
\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)
\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)
\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)
\(=-1\)
Bài giải
\(-3x\left|2020x+4040\right|=15\)
\(\left|2020x+4040\right|=15\text{ : }\left(-3x\right)\)
\(\left|2020\left(x+2\right)\right|=-5x\)
\(\Rightarrow\orbr{\begin{cases}2020\left(x+2\right)=5x\\2020\left(x+2\right)=-5x\end{cases}}\Rightarrow\orbr{\begin{cases}x+2=\frac{x}{404}\\x+2=\frac{x}{-404}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{x}{404}-x=x\left(\frac{1}{404}-1\right)=2\\\frac{x}{-404}-x=x\left(-\frac{1}{404}-1\right)=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\cdot\frac{-403}{404}=2\\x\cdot\frac{-405}{404}=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-808}{403}\\x=\frac{-808}{405}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{-808}{403}\text{ ; }\frac{-808}{405}\right\}\)