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\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(=\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)
\(=\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)
\(\frac{-x-18}{\left(x-1\right)\left(x-20\right)}=-\frac{3}{4}\)
\(\frac{-x-18}{x^2-21x+20}=\frac{-3}{4}\)
\(\frac{x+18}{x^2-21x+20}=\frac{3}{4}\)
\(4\left(x+18\right)=3\left(x^2-21x+20\right)\)
\(4x+72=3x^2-63x+60\)
\(3x^2-63x-4x=72-60\)
\(3x^2-67x=12\)
\(x\left(2x-67\right)=12\)
\(\Rightarrow x;2x-67\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Mà 2x - 67 lẻ.
Ta có bảng sau:
| 2x-67 | 2x | x | (2x - 67 ) . x |
| -3 | 64 | 32 | -96 ( loại) |
| -1 | 66 | 33 | -33 ( loại ) |
| 1 | 68 | 34 | 34 ( loại) |
| 3 | 70 | 35 | 105(loại) |
Do đó không có \(x\)thỏa mãn.
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\Leftrightarrow\dfrac{2}{x-3}-\dfrac{2}{x-2}+\dfrac{1}{x-8}-\dfrac{1}{x-3}+\dfrac{1}{x-20}-\dfrac{1}{x-8}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{2}{x-2}=\dfrac{-3}{4}\)
\(\Leftrightarrow4\left(x-2\right)-8\left(x-3\right)=-3\left(x-3\right)\left(x-2\right)\)
\(\Leftrightarrow4x-8-8x+24+3\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow3x^2-15x+18-4x+16=0\)
\(\Leftrightarrow3x^2-19x+34=0\)
\(\text{Δ}=\left(-19\right)^2-4\cdot3\cdot34=-47< 0\)
Do đó: Phương trình vô nghiệm
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
=> x = 12
a, 6/7 + (2/11 - 6/7) - (13/11 + 1)
= 6/7 + 2/11 - 6/7 - 13/11 - 1
= (6/7 - 6/7) - (13/11 - 2/11) - 1
= 0 - 1 - 1
= -2
Vì \(\left(3x-33\right)^{2016}\ge0;\left|y-7\right|\ge0\Leftrightarrow\left|y-7\right|^{2017}\ge0\)
=>\(\left(3x-33\right)^{2016}+\left|y-7\right|^{2017}\ge0\)
mà theo đề bài: \(\left(3x-33\right)^{2016}+\left|y-7\right|^{2017}\le0\)
=>\(\left(3x-33\right)^{2016}+\left|y-7\right|^{2017}=0\) <=>\(\left(3x-33\right)^{2016}=0;\left|y-7\right|^{2017}=0\)
- (3x-33)2016=0 <=> 3x-33=0 <=> 3x=33 <=> x=11
- |y-7|2017=0 <=> |y-7|=0 <=> y-7=0 <=> y=7
Vậy x=11 và y=7
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}+\dfrac{1}{x-20}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{x}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}+\dfrac{1}{x-20}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{x-1}=\dfrac{3}{4}\Rightarrow3x-3=4\Rightarrow x=\dfrac{7}{3}\)
Chúc bạn học tốt!
\(Ư\left(48\right)=\left\{1;-1;-2;2;3;-3;6;-6;8;-8;12;-12;-24;24;16;-16;48;-48\right\}\\ Ư\left(8\right)=\left\{1;-1;2;-2;-4;4;8;8\right\}\\ Ư\left(14\right)=\left\{-1;1;2;-2;7;-7;14;-14\right\}\\ Ư\left(20\right)=\left\{-1;1;2;-2;5;-5;4;-4;-10;10;-20;20\right\}\)