Question

Giải PT:

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

b/ \(\left(x-7\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)=72\)

Answer 1

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

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

\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)

\(\Leftrightarrow8x^2-66+100=0\)

\(\Leftrightarrow4x^2-33x+50=0\)

\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)

Answer 2

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

<=> (x²-9x+14)(x²-9x+20)=72

Đặt x²-9x+17=a

=> (a+3)(a-3)=72

<=> a²-9=72

<=> a²=81

=> a=\(\left\{9;-9\right\}\)

TH1: a=9

=> x²-9x+17=9

<=> x²-9x+8=0

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

=> x=\(\left\{1;8\right\}\)

TH2: a=-9

=> x²-9x+17=-9

<=> x²-9x+26=0

<=> x²-9x+20,25+5,75=0

<=> (x-4,5)²+5,75=0

=> x\(\in\varnothing\)

Vậy x=\(\left\{1;8\right\}\)