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\(\frac{1}{\left(x+1\right)^2\left(x+2\right)}=\frac{a}{x+1}+\frac{b}{\left(x+1\right)^2}+\frac{c}{x+2}\)
\(=\frac{a}{x+1}+\frac{b}{x+1^2}+\frac{c}{x+2}\)
\(=\frac{1}{\left(x+1\right)^2\left(x+2\right)=}=\frac{a}{\left(x+1\right)\left(x+2\right)}+\frac{b}{x+2}+\frac{c}{\left(x+1\right)^2\left(x+2\right)}\)
\(\frac{c}{\left(x+1\right)^2}+\frac{a}{\left(x+1\right)\left(x+2\right)}+\frac{b}{\left(x+2\right)}=1\)
\(=\frac{c}{x^2+2c+x+1}+\frac{a}{x^2+3a\left(x+2a\right)}+\frac{b}{x+2b}=1\)
\(=\frac{\left(c+a\right)}{x^2+\left(2+x+1+\frac{a}{x^2+3ax+2a}+\frac{b}{x+2b}\right)=1}\)
\(=\frac{c+a}{x^2+\left(2c+3a+b\right)}x+2a+2b=0\)
\(\frac{c+a=0}{2c+3b=0}2a+2b=0\)
\(c=b=-a\)
Vậy:.....
đơn giản
nhưng trả lời câu hỏi của tớ đã
a,\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
Ta có: \(x^2+5\ge0\) (vô lí)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-6\end{cases}}\)
Vậy ....
c, \(4x^2\left(x-1\right)-x+1=0\)
\(\Leftrightarrow4x^3-4x^2-x+1=0\)
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x^2-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x^2=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\frac{1}{2}\\x=1\end{cases}}\)
Vậy ....
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
ĐKXĐ: \(x\ne1,x\ne-3\)
PT đã cho \(\Leftrightarrow\frac{\left(x+2\right).\left(x-1\right)-\left(x+1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x-1\right)-\left(x+1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+x-2-x^2-4x-3=4\Leftrightarrow3x=-1\Leftrightarrow x=\frac{-1}{3}\)
Em kiểm tra lại đề bài nhé \(\frac{2}{x-y}\)hay \(\frac{2}{x-2}\)
dạ là x-y ạ