1, $(x-2)^{8}$ = $(x-2)^{6}$
2, $(x+2)^{2}$ = 36
3, $(x+2)^{3}$ = -27
1, $(x-2)^{8}$ = $(x-2)^{6}$ 2, $(x+2)^{2}$ = 36 3, $(x+2)^{3}$ = -27
By Skylar
By Skylar
1, $(x-2)^{8}$ = $(x-2)^{6}$
2, $(x+2)^{2}$ = 36
3, $(x+2)^{3}$ = -27
Đáp án:
1. Ta có :
`(x – 2)^8 = (x – 2)^6`
` <=> (x – 2)^8 – (x – 2)^6 = 0`
` <=> (x – 2)^6.[(x – 2)^2 – 1] = 0`
<=> \(\left[ \begin{array}{l}(x – 2)^6 = 0\\(x – 2)^2 – 1 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x – 2 = 0\\(x – 2)^2 = 1\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x=3 ; x = 1\end{array} \right.\)
Vậy `x = 1 ; x = 2 ; x = 3`
2. Ta có :
`(x + 2)^2 = 36`
<=> \(\left[ \begin{array}{l}x + 2 = 6\\x + 2 = -6\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
3. Ta có :
`(x + 2)^3 = -27`
` <=> (x + 2)^3 = (-3)^3`
` <=> x + 2 = -3`
` <=> x = -5`
Giải thích các bước giải:
`text{Đáp án:}`
`1.` \(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\)
`2.` \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
`3.` `x=-5`
`text{Giải thích các bước giải:}`
`1.`
`(x-2)^8=(x-2)^6`
`=>(x-2)^8-(x-2)^6=0`
`=>(x-2)^6.(x-2)^2-(x-2)^6=0`
`=>(x-2)^6.(x-2)^2-(x-2)^6 .1=0`
`=>(x-2)^6.[(x-2)^2-1]=0`
`=>`\(\left[ \begin{array}{l}(x-2)^6=0\\(x-2)^2 -1=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}(x-2)=0\\(x-2)^2=0+1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x-2=0\\(x-2)^2=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=0+2\\(x-2)^2=1^2\\(x-2)^2=(-1)^2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2\\x-2=1\\x-2=-1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2\\x=1+2\\x=-1+2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\)
`2.`
`(x+2)^2=36`
`=>`\(\left[ \begin{array}{l}(x+2)^2=6^2\\(x-2)^2=(-6)^2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x+2=6\\x+2=-6\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=6-2\\x=-6-2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
`3.`
`(x+2)^3=-27`
`=>(x+2)^3=(-3)^3`
`=>x+2=-3`
`=>x=-3-2`
`=>x=-5`