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AM
7 tháng 6 2015
ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)
\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)
\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)
\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)
\(\Leftrightarrow x^3+10x^2-32x-384=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)
\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)
DV
0
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
27 tháng 1
\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)
TH1: x<1
(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0
=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(5x+55=0\)
=>x=-11(nhận)
TH2: 1<=x<2
Phương trình (1) sẽ trở thành:
\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(7x+53=0\)
=>\(x=-\dfrac{53}{7}\left(loại\right)\)
TH3: 2<=x<3
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)
=>\(11x+45=0\)
=>\(x=-\dfrac{45}{11}\left(loại\right)\)
TH4: 3<=x<4
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)
=>\(-3x+27=0\)
=>x=9(loại)
TH5: 4<=x<5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)
=>\(25x-5=0\)
=>x=1/5(loại)
TH6: x>=5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)
=>35x-55=0
=>x=55/35(loại)
L
5 tháng 8 2019
\(\left(3x-4\right)\left(x-1\right)^3=0\)
\(\Rightarrow\orbr{\begin{cases}3x-4=0\\\left(x-1\right)^3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x=0+4\\x-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x=4\\x=0+1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\text{ }\left\{\frac{4}{3}\text{ ; }1\right\}\)
DD
18 tháng 5 2017
\(M\left(x\right)+N\left(x\right)\)
\(=5x^3-x^2-4+2x^4-2x^2+2x+1\)
\(=2x^4+5x^3-3x^2+2x-3\)
\(M\left(x\right)-N\left(x\right)\)
\(=5x^3-x^2-4-\left(2x^4-2x^2+2x+1\right)\)
\(=5x^3-x^2-4-2x^4+2x^2-2x-1\)
\(=-2x^4+5x^3+x^2-2x-5\)
\(M\left(x\right)+P\left(x\right)=N\left(x\right)\)
\(\Rightarrow P\left(x\right)=N\left(x\right)-M\left(x\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-\left(5x^3-x^2-4\right)\)
\(\Rightarrow P\left(x\right)=2x^4-2x^2+2x+1-5x^3+x^2+4\)
\(\Rightarrow P\left(x\right)=2x^4-5x^3-x^2+2x+5\)
NT
0
\(\left|x+4\right|=x-x^2\)
\(\Leftrightarrow\left|x+4\right|+x^2=x\)
Dễ thấy \(VT\ge0\Rightarrow VP\ge0\)hay \(x\ge0\)
\(\Rightarrow\left|x+4\right|=x+4\)
\(pt\Leftrightarrow x+4+x^2=x\)
\(\Leftrightarrow x^2+4=0\)
Mà \(x^2+4\ge0\)nên pt vô nghiệm