Ai giúp mình với !! (2+4+...2016+2018)/1019090= -3x2 — 4x
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Bài làm
2+4+...+2016+2018/1019090 = -3x² - 4x
Ta có: số số hạng tử của phân số 2+4+...+2016+2018/1019090 là:( 2018 - 2 ) : 2 + 1 = 1009 ( số hạng)
Tổng của tử đó là: ( 2018 + 2 ) . 1009 : 2 = 1019090
=> Ta được: 1019090/1019090 = -3x² - 4x
<=> -3x² - 4x = 1
<=> -3x² - 4x - 1 = 0
<=> -3x² - 3x - x - 1 = 0
<=> -3x( x + 1 ) -( x + 1 ) = 0
<=> ( x + 1 )( -3x - 1 ) = 0
<=> x + 1 = 0 hoặc -3x - 1 = 0
<=> x = -1 hoặc x = 1/-3
Vậy nghiệm phương trình là: S = { -1; -1/3 }
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\(\left(3x-5\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)
\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)
\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)
\(\Leftrightarrow3x^2+2+7x=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)
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Đặt A=1+42+44+46+48+...+42016+42018
42A=42+44+46+48+...+42018+42020
42A-A=42020-42018
15A=42020-22018
A=42020-42018/15
Tìm x biết ( 2 cách )
( x - 2 ) mũ 2018 = ( x - 2 ) mũ 2016
Ai giúp mình với mai mình phải nộp bài rồi
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\(\left(x-2\right)^{2018}=\left(x-2\right)^{2016}\)
=> \(\left(x-2\right)^{2018}-\left(x-2\right)^{2016}=0\)
=> \(\left(x-2\right)^{2016}\left(\left(x-2\right)^2-1\right)=0\)
=> \(\orbr{\begin{cases}\left(x-2\right)^{2016}=0\\\left(\left(x-2\right)^2-1\right)=0\end{cases}}\)
=> x = 2
và (x - 2)2 = 1 => x - 2 = 1 ; -1
=> x = 3 ; x = 1
=> x = 1 ; 2 ; 3
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2+3+...+2018=(1+2018)+(2+2017)+...(1009+1010)=2019x1009=2037171
B=(1+2019)+(3+2017)+...+(1009+1011)=2020x505=1020100
C=(2020+2)+(2018+4)+...+(1010+1012)=2022x505=1021110
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\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\frac{2016-1}{2016}+\frac{2017-1}{2017}+\frac{2018-1}{2018}+\frac{2015+3}{2015}\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(=4+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2015}-\frac{1}{2018}\)
mà \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2017};\frac{1}{2018}\)
\(\Rightarrow A>4\)
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a) \(\left(x^3+3x^2-8x-20\right)\div\left(x+2\right)\)
\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(10x+20\right)\right]\div\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x-10\right)\div\left(x+2\right)\)
\(=x^2+x-10\) \(\left(x\ne-2\right)\)
b,c bn tự đặt chia
\(\frac{2+4+...+2016+2018}{1019090}=-3x^2-4x.\)
\(\Leftrightarrow\frac{1009\left(1009+1\right)}{1019090}=-3x^2-4x\)
\(\Leftrightarrow\frac{1009.1010}{1009.1010}=-3x^2-4x\)
\(\Leftrightarrow3x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2-x^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{3}\end{cases}}}\)
Vậy....