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![](https://rs.olm.vn/images/avt/0.png?1311)
\(2+23x=0\)
<=> \(23x=-2\)
<=> \(x=-\frac{2}{23}\)
KL: \(x=-\frac{2}{23}\) là nghiệm của đa thức
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi 3 số liên tiếp là (a-1);a;(a+1);
Ta có (a-1)a+(a-1)(a+1)+(a+1)a =362
=>a^2-a+a^2-1+a^2+a=362
=>3a^2=363
=>a^2=121
=> a=11
=> Ba số tự nhiên liên tiếp cần tìm là :10;11;12
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=0-1=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: 2/3x-2/3=-1
=>2/3x=-1/3
hay x=-1/2
b: \(\dfrac{x-1}{3}=\dfrac{-1}{2}\)
=>x-1=-3/2
hay x=-1/2
c: |x+2/5|-2=-1/4
=>|x+2/5|=7/4
=>x+2/5=7/4 hoặc x+2/5=-7/4
=>x=27/20 hoặc x=-43/20
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\) (ĐK: \(x\ne\frac{3}{2};x\ne-2\))
\(\Rightarrow\left(x+2\right)\left(5x-4\right)=\left(7+4x\right)\left(3-2x\right)\)
\(\Rightarrow5x^2-4x+10x-8=21-14x+12x-8x^2\)
\(\Rightarrow13x^2+8x-29=0\)
\(\Rightarrow13\left(x^2+\frac{8}{13}x-\frac{29}{13}\right)=0\)
\(\Rightarrow13\left[x^2+2.\frac{4}{13}.x+\left(\frac{4}{13}\right)^2-\left(\frac{4}{13}\right)^2-\frac{29}{13}\right]=0\)
\(\Rightarrow13\left[\left(x+\frac{4}{13}\right)^2-\frac{393}{169}\right]=0\)
\(\Rightarrow13\left(x+\frac{4}{13}\right)^2-\frac{393}{13}=0\)
\(\Rightarrow\left(x+\frac{4}{13}\right)^2=\frac{393}{169}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{13}=\sqrt{\frac{393}{169}}=\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4+\sqrt{393}}{13}\\x+\frac{4}{3}=-\sqrt{\frac{393}{169}}=-\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4-\sqrt{393}}{13}\end{cases}}\)
Vậy biểu thức có 2 nghiệm \(x=\left\{\frac{-4+\sqrt{393}}{13};\frac{-4-\sqrt{393}}{13}\right\}\)
b/ \(\frac{x-1}{99}+\frac{x-2}{98}-\frac{x-3}{97}-\frac{x-4}{96}=0\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1-\left(\frac{x-3}{97}-1\right)-\left(\frac{x-4}{96}-1\right)=0\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
=> x - 100 = 0 => x = 100
Vậy x = 100
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì x chia 8;10;15;20 đều dư 3
=> x-3 chia hết cho 8;10;15;20
=> x-3\(\in\) BC(8;10;15;20) = {0;240;480;...}
=> x \(\in\){3;243;483;...}
Mà x từ trong khoảng 230-300 => x = 243
![](https://rs.olm.vn/images/avt/0.png?1311)