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\(\left(\dfrac{1}{3}x^3y\right)\left(-xy\right)^2=\left(\dfrac{1}{3}x^3y\right).x^2y^2=\dfrac{1}{3}x^5y^3\)
Bậc: 5
Hệ số: \(\dfrac{1}{3}\)
ta có (sau khi nhân) đơn thưc
-\(\frac{1}{3}\)x5y3
bậc là 5. hệ số -1/3
a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31+61+101+151+...+x.(x−1)2=20092007
26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62+122+202+302+...+(x−1).x2=20092007
22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32+3.42+4.52+5.62+...+(x−1).x2=20092007
2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−31+31−41+41−51+51−61+...+x−11−x1)=20092007
2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−x1)=20092007
1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2=20092007
2x=22009\frac{2}{x}=\frac{2}{2009}x2=20092
=> x = 2009
Ta có:
\(\frac{-2105}{2016}< -1\)
\(\frac{-2001}{2000}>-1\)
\(\Rightarrow\frac{-2105}{2016}< -1< \frac{-2001}{2000}\Rightarrow\frac{-2015}{2016}< \frac{-2001}{2000}\)
Ta có:
\(1-\frac{-999}{605}=\frac{1640}{605}\)
\(1-\frac{1199}{-805}=\frac{1640}{805}\)
Vì \(\frac{1640}{605}< \frac{1640}{805}\) nên \(\frac{-999}{605}>\frac{1199}{-805}\)
-2015/2016 và -2001/2000
Ta có: 2015/2016 < 1
2001/2000 > 1
=> 2015/2016 < 2001/200 => -2015/2016 > -2001/2000
\(P=\left|x-28\right|+\left|x-3\right|+\left|x-2020\right|\)
\(=\left(\left|x-3\right|+\left|x-2020\right|\right)+\left|x-28\right|\)
Đặt \(A=\left|x-3\right|+\left|x-2020\right|\)
Ta có: \(A=\left|x-3\right|+\left|x-2020\right|\)
\(=\left|x-3\right|+\left|2020-x\right|\ge\left|x-3+2020-x\right|=2017\left(1\right)\)
Dấu"="xảy ra \(\Leftrightarrow\left(x-3\right)\left(2020-x\right)\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-3\ge0\\2020-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-3< 0\\2020-x< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le2020\end{cases}}\)hoặc \(\hept{\begin{cases}x< 3\\x>2020\end{cases}\left(loai\right)}\)
\(\Leftrightarrow3\le x\le2020\)
Ta có: \(\left|x-28\right|\ge0;\forall x\left(2\right)\)
Dấu"="xảy ra \(\Leftrightarrow\left|x-28\right|=0\)
\(\Leftrightarrow x=28\)
Từ (1) và (2)\(\Rightarrow A+\left|x-28\right|\ge2017\)
Hay \(P\ge2017\)
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}3\le x\le2020\\x=28\end{cases}}\Leftrightarrow x=28\)
Vậy \(P_{min}=2017\Leftrightarrow x=28\)
Ta có:\(\frac{3-x}{2021}+\frac{2020-x}{2019}+\frac{4033-x}{2017}+\frac{6042-x}{2015}=10\)
\(\Leftrightarrow\frac{3-x}{2021}-1+\frac{2020-x}{2019}-2+\frac{4033-x}{2017}-3+\frac{6042-x}{2015}-4=0\)
\(\Leftrightarrow\frac{3-x-2021}{2021}+\frac{2020-x-4038}{2019}+\frac{4033-x-6051}{2017}+\frac{6042-x-8060}{2015}=0\)
\(\Leftrightarrow\frac{-2018-x}{2021}+\frac{-2018-x}{2019}+\frac{-2018-x}{2017}+\frac{-2018-x}{2015}=0\)
\(\Leftrightarrow-\left(2018+x\right)\left(\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}\right)=0\)
\(\Leftrightarrow2018+x=0.Do\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}>0\)
\(\Leftrightarrow x=-2018\)
V...
a/ \(A.B=\left(2^{2015}.7^{2020}\right).\left(2^{2020}.7^{2015}\right)\)
\(A.B=\left(2^{2015}.2^{2020}\right).\left(7^{2020}.7^{2015}\right)\)
\(A.B=2^{4035}.7^{4035}\)
Còn nhân 2 đơn thức thì giống như trên :
\(\left(x^{2015}.y^{2020}\right).\left(x^{2020}.y^{2015}\right)\)
\(=\left(x^{2015}.x^{2020}\right).\left(y^{2020}.y^{2015}\right)\)
\(=x^{4035}.y^{4035}\)
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