a,(x+2)(2x-5)=x ²+4x+4 b,x ²-16=0 c,x ^4-16=0 d.x ²+5x+6=0 e,x ²-8x+12=0 f,x^4=1 giúp iem vs mơn nhìu 31/10/2021 Bởi Josie a,(x+2)(2x-5)=x ²+4x+4 b,x ²-16=0 c,x ^4-16=0 d.x ²+5x+6=0 e,x ²-8x+12=0 f,x^4=1 giúp iem vs mơn nhìu
Đáp án + Giải thích các bước giải: `a,(x+2)(2x-5)=x^2+4x+4` `⇔2x^2-5x+4x-10-x^2-4x-4=0` `⇔x^2-5x-14=0` `⇔(x^2+2x)-(7x+14)=0` `⇔x(x+2)-7(x+2)=0` `⇔(x+2)(x-7)=0` `⇔` \(\left[ \begin{array}{l}x+2=0\\x-7=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-2\\x=7\end{array} \right.\) Vậy `x∈{-2;7}` `b,x^2-16=0` `⇔(x-4)(x+4)=0` `⇔` \(\left[ \begin{array}{l}x-4=0\\x+4=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\) Vậy `x∈{4;-4}` `c,x^4-16=0` `⇔(x^2-4)(x^2+4)=0` `⇔(x^2-4)=0` `[ Do (x^2+4) > 0 ]` `⇔(x-2)(x+2)=0` `⇔` \(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) Vậy `x∈{2;-2}` `d,x^2+5x+6=0` `⇔(x^2+2x)+(3x+6)=0` `⇔x(x+2)+3(x+2)=0` `⇔(x+2)(x+3)=0` `⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\) Vậy `x∈{-2;-3}` `e,x^2-8x+12=0` `⇔(x^2-6x)-(2x-12)=0` `⇔x(x-6)-2(x-6)=0` `⇔(x-6)(x-2)=0` `⇔` \(\left[ \begin{array}{l}x-6=0\\x-2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=6\\x=2\end{array} \right.\) Vậy `x∈{6;2}` `f,x^4=1` `⇔` \(\left[ \begin{array}{l}x^4=1^4\\x^4=(-1)^4\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) Vậy `x∈{1;-1}` Bình luận
$#Dinosaurss$ a, `(x+2)(2x-5)=x^(2)+4x+4` `⇔(x+2)(2x-5)=(x+2)^2` `⇔(x+2)(2x-5)-(x+2)^2=0` `⇔(x+2)(2x-5-x-2)=0` `⇔(x+2)(x-7)=0` `⇔x+2=0` `x-7=0` `⇔x=-2` `x=7` Vậy `S={-2;7}` b, `x^(2)-16=0` `⇔x^(2)=16` `⇔x=±4` Vậy `S={±4}` c,` x^(4)-16=0` `⇔x^4=16` `⇔x=±2` Vậy `S={±2}` d, `x^(2)+5x+6=0` `⇔x(2)+2x+3x+6=0` `⇔x(x+2)+3(x+2)=0` `⇔(x+2)(x+3)=0` `⇔x+2=0` `x+3=0` `⇔x=-2` `x=-3` Vậy `S={-2;-3}` e,`x^(2)-8x+12=0` `⇔x^(2)-6x-2x+12=0` `⇔x(x-6)-2(x-6)=0` `⇔(x-6)(x-2)=0` `⇔x-6=0` `x-2=0` `⇔x=6` `x=2` Vậy `S={6;2}` f, `x^4=1` `⇔x^4=1^4` `⇔x=±1` Vậy `S={±1}` @DuongQuynhChi<3 Bình luận
Đáp án + Giải thích các bước giải:
`a,(x+2)(2x-5)=x^2+4x+4`
`⇔2x^2-5x+4x-10-x^2-4x-4=0`
`⇔x^2-5x-14=0`
`⇔(x^2+2x)-(7x+14)=0`
`⇔x(x+2)-7(x+2)=0`
`⇔(x+2)(x-7)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x-7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=7\end{array} \right.\)
Vậy `x∈{-2;7}`
`b,x^2-16=0`
`⇔(x-4)(x+4)=0`
`⇔` \(\left[ \begin{array}{l}x-4=0\\x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy `x∈{4;-4}`
`c,x^4-16=0`
`⇔(x^2-4)(x^2+4)=0`
`⇔(x^2-4)=0` `[ Do (x^2+4) > 0 ]`
`⇔(x-2)(x+2)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `x∈{2;-2}`
`d,x^2+5x+6=0`
`⇔(x^2+2x)+(3x+6)=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(x+2)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy `x∈{-2;-3}`
`e,x^2-8x+12=0`
`⇔(x^2-6x)-(2x-12)=0`
`⇔x(x-6)-2(x-6)=0`
`⇔(x-6)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x-6=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=6\\x=2\end{array} \right.\)
Vậy `x∈{6;2}`
`f,x^4=1`
`⇔` \(\left[ \begin{array}{l}x^4=1^4\\x^4=(-1)^4\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `x∈{1;-1}`
$#Dinosaurss$
a, `(x+2)(2x-5)=x^(2)+4x+4`
`⇔(x+2)(2x-5)=(x+2)^2`
`⇔(x+2)(2x-5)-(x+2)^2=0`
`⇔(x+2)(2x-5-x-2)=0`
`⇔(x+2)(x-7)=0`
`⇔x+2=0`
`x-7=0`
`⇔x=-2`
`x=7`
Vậy `S={-2;7}`
b, `x^(2)-16=0`
`⇔x^(2)=16`
`⇔x=±4`
Vậy `S={±4}`
c,` x^(4)-16=0`
`⇔x^4=16`
`⇔x=±2`
Vậy `S={±2}`
d, `x^(2)+5x+6=0`
`⇔x(2)+2x+3x+6=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(x+2)(x+3)=0`
`⇔x+2=0`
`x+3=0`
`⇔x=-2`
`x=-3`
Vậy `S={-2;-3}`
e,`x^(2)-8x+12=0`
`⇔x^(2)-6x-2x+12=0`
`⇔x(x-6)-2(x-6)=0`
`⇔(x-6)(x-2)=0`
`⇔x-6=0`
`x-2=0`
`⇔x=6`
`x=2`
Vậy `S={6;2}`
f, `x^4=1`
`⇔x^4=1^4`
`⇔x=±1`
Vậy `S={±1}`
@DuongQuynhChi<3