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\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
\(ĐKXĐ:x\ne3;x\ne-1\)
Nếu x=0 là nghiệm của phương trình
Nếu x khác 0 ta có:
\(\frac{1}{2\left(x-3\right)}+\frac{1}{2\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x-1+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{2x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow2x-4=4\)
\(\Leftrightarrow x=4\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)
<=> \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x2-6x=0
<=> 2x(x-3)=0
<=> \(\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
ĐCĐK x khác -1 và x khác 3 => x=0
Vậy x=0 là nghiệm của phương trình
Lời giải:
a)
\(\frac{1}{(1-x)(2-x)}+\frac{2}{(2-x)(3-x)}+\frac{3}{(1-x)(x-3)}=\frac{1}{(x-1)(x-2)}+\frac{2}{(x-2)(x-3)}-\frac{3}{(x-1)(x-3)}\)
\(=\frac{x-3}{(x-1)(x-2)(x-3)}+\frac{2(x-1)}{(x-1)(x-2)(x-3)}-\frac{3(x-2)}{(x-1)(x-2)(x-3)}\)
\(=\frac{x-3+2(x-1)-3(x-2)}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)(x-2)(x-3)}\)
b)
\(\frac{x^2}{x+1}+\frac{2x}{x^2-1}-\frac{1}{1-x}+1=\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x-1}+1\)
\(=\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{x}{x-1}=\frac{x^2(x-1)}{(x+1)(x-1)}+\frac{2x}{(x-1)(x+1)}+\frac{x(x+1)}{(x-1)(x+1)}\)
\(=\frac{x^3+3x}{(x-1)(x+1)}=\frac{x^3+3x}{x^2-1}\)
c)
\(\frac{1}{x^3-x}-\frac{1}{x(x-1)}+\frac{2}{x^2-1}=\frac{1}{x(x-1)(x+1)}-\frac{x+1}{x(x-1)(x+1)}+\frac{2x}{x(x-1)(x+1)}\)
\(=\frac{x}{x(x-1)(x+1)}=\frac{1}{(x-1)(x+1)}=\frac{1}{x^2-1}\)
ta có
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
=> ĐK: \(x\ne\left\{0;-1;-2;...;-99;-100\right\}\)
Đây là dạng dãy số đặc biệt, bạn có thể giải như sau:
Ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{100}{x^2+100x}=\frac{100}{101}\)
\(\Leftrightarrow x^2+100x=101\)
\(\Leftrightarrow x^2+100x-101=0\)
\(\Leftrightarrow x^2+101x-x-101=0\)
\(\Leftrightarrow x\left(x+101\right)-\left(x+101\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+101\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+101=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(n\right)\\x=-101\left(n\right)\end{cases}}\)
Vậy: S={1;-101)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+9}-\frac{1}{x+10}\)
\(=\frac{1}{x}-\frac{1}{x+10}\)
\(=\frac{10}{x\left(x+10\right)}\)
Bạn tham khảo nhé!
Phân thức cuối hình như mẫu sai rồi bạn
Phải là (x+9)(x+10) mới đúng chứ
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+.....+\frac{1}{x+99}-\frac{1}{x+100}\)
\(=\frac{1}{x}-\frac{1}{x+100}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{x\left(x+100\right)}\)