Đáp án:
Giải thích các bước giải:
Bài 7 :
a, `(x^2-x)^2-4(x^2-x)+4=0`
`<=>(x^2-x)^2-2.(x^2-x).2+2^2=0`
`<=>(x^2-x-2)^2=0`
`<=>x^2-x-2=0`
`<=>x^2+x-2x-2=0`
`<=>x(x+1)-2(x+1)=0`
`<=>(x-2)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
b, `(2x+1)^2-2x-1=2`
`<=>4x^2+4x+1-2x-1-2=0`
`<=>4x^2+2x-2=0`
`<=>2x^2+x-1=0`
`<=>2x^2+2x-x-1=0`
`<=>2x(x+1)-(x+1)=0`
`<=>(x+1)(2x-1)=0`
`<=>` \(\left[ \begin{array}{l}2x-1=0\\x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-1\end{array} \right.\)
c, `(x^2-3x)^2+5(x^2-3x)+6=0`
Đặt `x^2-3x=a`
Ta được :
`a^2+5a+6=0`
`<=>a^2+2a+3a+6=0`
`<=>a(a+2)+3(a+2)=0`
`<=>(a+3)(a+2)=0`
`<=>` \(\left[ \begin{array}{l}a+3=0\\a+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}a=-3\\a=-2\end{array} \right.\)
Suy ra :
Với `x^2-3x=-3`
`<=>x^2-3x+3=0`
`<=>x^2-2.x.3/2+9/4+3/4=0`
`<=>(x+3/2)^2=-3/4` ( vô lí )
`<=>x \in ` $\varnothing$
Với `x^2-3x=-2`
`<=>x^2-3x+2=0`
`<=>x^2-x-2x+2=0`
`<=>x(x-1)-2(x-1)=0`
`<=>(x-2)(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
d, `(x^2-x-1)(x^2-x)-2=0`
Đặt `x^2-x=a`
Ta được :
`(a-1).a-2=0`
`<=>a^2-a-2=0`
`<=>a^2+a-2a-2=0`
`<=>a(a+1)-2(a+1)=0`
`<=>(a-2)(a+1)=0`
`<=>` $\begin{cases}a-2=0\\a+1=0\end{cases}$
`<=>` $\begin{cases}a=2\\a=-1\end{cases}$
Với `x^2-x=2`
`<=>x^2-x-2=0`
`<=>x^2+x-2x-2=0`
`<=>x(x+1)-2(x+1)=0`
`<=>(x-2)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Với `x^2-x=-1`
`<=>x^2-x+1=0`
`<=>x^2-2.x.1/2+1/4+3/4=0`
`<=>(x-1/2)^2=-3/4` ( vô lí )
`<=>x \in ` $\varnothing$
Bài 8 :
Thay `t=3` vào phương trình , ta có :
`2/(5-3)-a-3=2a(a+2)`
`<=>1-a-3-2a^2-4a=0`
`<=>-2a^2-5a-2=0`
`<=>2a^2+5a+2=0`
`<=>2a^2+4a+a+2=0`
`<=>2a(a+2)+(a+2)=0`
`<=>(2a+1)(a+2)=0`
`<=>` \(\left[ \begin{array}{l}2a+1=0\\a+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\a=-2\end{array} \right.\)
