Câu hỏi của Kim Ngưu

Question

Giải pt sau:

1) (9x2-4)(x+1)=(3x+2)(x2-1)

2) (x-1)2-1+x2=(1-x)(x+3)

3) x2-7x+6=0

4) x4 -4x3+12-9=0

5) (x2+x)2+4(x2+x)-12=0

6) x(x+1)(x2+x+1)=42

7) (x+2)(x-2)(x2-10)=72

8) (x-1)(x-3)(x+5)(x+7)-...

Duy Đỗ Ngọc Tuấn · Accepted Answer

3) $x^2-7x+6=0$

$\Leftrightarrow x^2-6x-x+6=0$

$\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0$

$\Leftrightarrow\left(x-6\right)\left(x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-6=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\x=1\end{matrix}\right.$

S=$\left\{6;1\right\}$

Câu trả lời: 3) $x^2-7x+6=0$

$\Leftrightarrow x^2-6x-x+6=0$

$\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0$

$\Leftrightarrow\left(x-6\right)\left(x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-6=0\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\x=1\end{matrix}\right.$

S=$\left\{6;1\right\}$

\(\frac{90}{x}-\frac{36}{x-6}=2\)	\(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)	\(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
\(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)	\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)	\(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)

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