Question

[Trí Phạm]

Giải PT:

a/ \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

b/ (x−7)(x−2)(x−4)(x−5)=72

Answer 1

[Kiệt Nguyễn]

\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)

\(\Rightarrow2x^2-4x=10x^2-70x+100\)

\(\Rightarrow8x^2-66x+100=0\)

Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)

Answer 2

[Edogawa Conan]

a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)

<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(2x=10\left(x-5\right)\)

<=> 2x - 10x = -50

<=> -8x = -50

<=>x = 6,25

Vậy S = {6,25}

b) (x - 7)(x - 2)(x - 4)(x - 5) = 72

<=> (x² - 9x + 14)(x² - 9x + 20) = 72

Đặt x² - 9x + 14 = t <=> t(t + 6) = 72

<=> t² + 6t - 72 = 0

<=> t² + 12t - 6t - 72 = 0

<=> (t + 12)(t - 6) = 0

<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)

<=> (x - 1)(x - 8) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Vậy S = {1; 8}