Đáp án+Giải thích các bước giải:
`d) 2((x^2+x+1)/(x-1))^2 -7 = (13(x^3-1))/((x-1)^2) (ĐKXĐ: x \ne 1)`
`<=> 2((x^2+x+1)/(x-1))^2 -7 = (13(x^3-1))/((x-1)^2)`
`<=> (2(x^2+x+1)^2)/((x-1)^2)-(7(x-1)^2)/((x-1)^2)= (13(x^3-1))/((x-1)^2)`
`=> 2(x^4+x^2+1+2x^3+2x^2+2x)-7(x^2-2x+1)=13x^3-13`
`<=>2(x^4+2x^3+3x^2+2x+1)-7(x^2-2x+1)=13x^3-13`
`<=> 2x^4+4x^3+6x^2+4x+2 -7x^2+14x-7-13x^3+13=0`
`<=> 2x^4 -9x^3-x^2+18x+8=0`
`<=>(x-4)(x-2)(x+1)(2x+1)=0`
`<=>`\(\left[ \begin{array}{l}x-4=0\\x-2=0\\x+1=0\\2x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=2\\x=-1\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{4; 2; -1; -1/2}`
`b) (x^2-2x+1)/(x^2-2x+2)+(x^2-2x+2)/(x^2-2x+3)=7/6 (1)`
Đặt `x^2-2x+2=t(2)`. Khi đó phương trình `(1)` trở thành:
`(t-1)/t+t/(t+1) = 7/6`
`<=>(6(t^2-1))/(6t(t+1))+(6(t^2))/(6t(t+1))=(7t(t+1))/(6t(t+1))`
`=> 6t^2-6+6t^2=7t^2+7t`
`<=> 5t^2-7t-6=0`
`<=>(t-2)(5t+3)=0`
`<=>` \(\left[ \begin{array}{l}t-2=0\\5t+3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}t=2\\t=-\dfrac{3}{5}\end{array} \right.\)
`+)` Thay `t=2` vao `(2)` ta được:
`x^-2x+2=2`
`<=> x^-2x=0`
`<=> x(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
`+)` Thay `t=-3/5` vao `(2)` ta được:
`x^2-2x+2=-3/5`
`<=> (x-1)^2+1=-3/5`
`<=> (x-1)^2 = -8/5`(vô nghiệm)
Vậy `x∈0; 2}`
