Đáp án:1)\(\left[ \begin{array}{l}x=3\\x=\frac{8}{3}\end{array} \right.\)
2)$2(-18x^{2}+3x+13)=0$
3)Vô lí
Giải thích các bước giải:
1)$\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1$ (đk: $x\neq2; x\neq4$)
⇔$\frac{x-3}{x-2}+\frac{x-2}{x-4}+1=0$
⇒$(x-3)(x-4)+(x-2)^{2}+(x-2)(x-4)=0$
⇔$x^{2}-7x+12+x^{2}-4x+4+x^{2}-6x+8=0$
⇔$3x^{2}-17x+24=0⇔ (x-3)(x-\frac{8}{3})=0$
⇔\(\left[ \begin{array}{l}x-3=0\\x-\frac{8}{3}=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=3\\x=\frac{8}{3}\end{array} \right.\)
2)$\frac{5x-2}{12}-2x^{2}+1=\frac{x-3}{6}+\frac{1-x^{2}}{4}$ (MC:24)
⇒$2(5x-2)-24·2x^{2}+24-4(x-3)-6(1-x^{2})=0$
⇔$10x-4-48x^{2}+24-4x+12-6+6x^{2}=0$
⇔$-36x^{2}+6x+26=0⇔ 2(-18x^{2}+3x+13)=0$
3)$\frac{14x-3}{15}=\frac{x+1}{9}+\frac{37x-23}{45}$ (MC:45)
⇔$\frac{14x-3}{15}-\frac{x+1}{9}-\frac{37x-23}{45}=0$
⇒$3(14x-3)-5(x+1)-37x+23=0$
⇔$42x-9-5x-5-37x+23=0⇔(42x-5x-37x)=-23+5+9⇔ 0x=-9$ (Vô lí)
