giúp em với ạ a)(5-x)(5+x)-(2x-1)^2
b)(x^4+2x^3+10x-25):(x^2+5)
c)(x+3)(x^2+3x-5)-x(x-2)^2
tìm x d):a)x^4-8x^2-9=0
e)6^2-(2x+5)(3x-2)=-12
giúp em với ạ a)(5-x)(5+x)-(2x-1)^2 b)(x^4+2x^3+10x-25):(x^2+5) c)(x+3)(x^2+3x-5)-x(x-2)^2 tìm x d):a)x^4-8x^2-9=0 e)6^2-(2x+5)(3x-2)=-12
By Madeline
Đáp án:
Giải thích các bước giải:
a) $(5-x)(5+x)-(2x-1)^{2}$
= $25-x^{2}-(4x^{2}-4x+1)$
= $5x^{2}+4x+24$
b) $(x^{4}+2x^{3}+10x-25):(x^{2}+5)$
= $[(x^{2}-5)(x^{2}+5)+2x(x^{2}+5)]:(x^{2}+5)$
= $(x^{2}+5)(x^{2}-5+2x):(x^{2}+5)$
= $x^{2}+2x-5$
c) $(x+3)(x^{2}+3x-5)-x(x-2)^{2}$
= $x^{3}+6x^{2}+4x-15-x(x^{2}-4x+4)$
= $10x^{2}+8x-15$
d) $x^{4}-8x^{2}-9=0$
⇔ $x^{4}+x^{2}-9x^{2}-9=0$
⇔ $x^{2}(x^{2}+1)-9(x^{2}+1)=0$
⇔ $(x^{2}+1)(x^{2}-9)=0$
⇔ $x^{2}-9=0$
⇔ $(x-3)(x+3)=0$
⇔ $x=±3$
e) $6x^{2}-(2x+5)(3x-2)=-12$
⇔ $6x^{2}-6x^{2}-11x+10=-12$
⇔ $-11x=-22$
⇔ $x=2$
Bài `1`:
`a) (5 – x)(5 + x) – (2x – 1)^2`
`= 5^2 – x^2 – [(2x)^2 – 2. 2x + 1^2]`
`= 25 – x^2 – (4x^2 – 4x + 1)`
`= 25 – x^2 – 4x^2 + 4x – 1`
`= -5x^2 + 4x + 24`
`b) (x^4 + 2x^3 + 10x – 25) : (x^2 + 5)`
`= [(x^4 – 25) + (2x^3 + 10x)] : (x^2 + 5)`
`= {[(x^2)^2 – 5^2] + 2x(x^2 + 5)} : (x^2 + 5)`
`= [(x^2 + 5)(x^2 – 5) + 2x(x^2 + 5)] : (x^2 + 5)`
`= (x^2 + 5)(x^2 – 5 + 2x) : (x^2 + 5)`
`= x^2 – 5 + 2x`
Bài `2`:
`a) x^4 – 8x^2 – 9 = 0`
`=> (x^2)^2 – 2x^2. 4 + 16 – 25 = 0`
`=> (x^2)^2 – 2x^2. 4 + 4^2 = 25`
`=> (x^2 – 4)^2 = (+-5)^2`
`=>` \(\left\{\begin{matrix}x^2 – 4=5\\x^2 – 4 = -5\end{matrix}\right.\)
`=>` \(\left\{\begin{matrix}x^2 =9\\x^2 =-1\end{matrix}\right.\)
`=>` \(\left\{\begin{matrix}x=\pm3\\x \in\emptyset \end{matrix}\right.\)
Vậy `x = +-3`
`b) 6x^2 – (2x + 5)(3x – 2) = -12`
`=> 6x^2 – (2x. 3x – 2x. 2 + 5. 3x – 5. 2) = -12`
`=> 6x^2 – (6x^2 – 4x + 15x – 10) = -12`
`=> 6x^2 – 6x^2 + 4x – 15x + 10 = -12`
`=> -11x + 10 = -12`
`=> -11x = -22`
`=> x = 2`
Vậy `x = 2`