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a. ( x - 15). ( x + 4 ) = 0
\(\Rightarrow\)( x - 15)=0 hoặc ( x + 4 ) =0
x = 0 + 15 x = 0 -4
x = 15 x = -4
Vậy x = 15 x = -4
\(x^3+x^2y-2x^2-xy-y^2+3y+x+2006\)
\(=x^2\left(x+y-2\right)-y\left(x+y-2\right)+y+x-2+2004\)
= 2004
B = 3 + 32 + 33 + ... + 32006
2B + 3 = 3B - B + 3 = (32 + 33 + 34 + ... + 32007) - (3 + 32 + 33 + ... + 32006) + 3 = 32007 - 3 + 3 = 32007 = 3x => x = 2007
\(B=3+3^2+3^3+.....+3^{2006}\)
\(\Rightarrow3B=3^2+3^3+3^4+.....+3^{2007}\)
\(\Rightarrow2B=3^{2007}-3\)
\(\Rightarrow B=\frac{3^{2007}-3}{2}\)
\(2B+3=3^x\)
\(\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\)
\(\Rightarrow3^{2007}-3+3=3^x\)
\(\Rightarrow3^{2007}=3^x\)
\(\Rightarrow x=2007\)
a: =>3/2x=5/6
hay x=5/6:3/2=5/6x2/3=10/18=5/9
b: \(\Leftrightarrow219\cdot1-7\left(x+1\right)=100\)
=>7(x+1)=119
=>x+1=17
hay x=16
kết quả là 2008 đấy bạn
nếu nhà bạn có máy tính thì chỉ cần bấm phương trình x thì sẽ ra kết quả thôi
\(\frac{x-1}{2007}+\frac{x-2}{2006}+\frac{x-3}{2005}=\frac{x-4}{2004}+\frac{x-5}{2003}+\frac{x-6}{2002}\)
=> \(\left(\frac{x-1}{2007}-1\right)+\left(\frac{x-2}{2006}-1\right)+\left(\frac{x-3}{2005}-1\right)=\left(\frac{x-4}{2004}-1\right)+\left(\frac{x-5}{2003}-1\right)+\left(\frac{x-6}{2002}-1\right)\)
=> \(\frac{x-1+2007}{2007}+\frac{x-2+2006}{2006}+\frac{x-3+2005}{2005}=\frac{x-4+2004}{2004}+\frac{x-5+2003}{2003}+\frac{x-6+2002}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}=\frac{x-2008}{2004}+\frac{x-2008}{2003}+\frac{x-2008}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}-\frac{x-2008}{2004}-\frac{x-2008}{2003}-\frac{x-2008}{2002}=0\)
=> \(\left(x-2008\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2008 = 0 => x = 2008
Vậy x = 2008
a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
x2-2005x-2006 = 0
x2+1x -2006x-2006=0
x(x+1)-2006(x+1)=0
(x-20016)(x+1)=0
=> x=2006 hoặc =-1