Toán Lớp 8: (x-2)^2+(3-2x)^2-(4x-x)(x-5(=(x-3)^2
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: (x-2)^2+(3-2x)^2-(4x-x)(x-5(=(x-3)^2
Comments ( 2 )
Giải đáp:
S={-1;-4}
Lời giải và giải thích chi tiết:
(x-2)^2+(3-2x)^2-(4x-x)(x-5)=(x-3)^2
<=>x^2-4x+4+9-12x+4x^2-3x(x-5)=x^2-6x+9
<=>5x^2-16x+13-3x^2+15x-x^2+6x-9=0
<=>x^2+5x+4=0
<=>x^2+x+4x+4=0
<=>x(x+1)+4(x+1)=0
<=>(x+1)(x+4)=0
<=>[(x+1=0),(x+4=0):}
<=>[(x=-1),(x=-4):}
S={-1;-4}
#kkk
$\left(x-2\right)^2+\left(3-2x\right)^2-\left(4x-x\right)\left(x-5\right)=\left(x-3\right)^2$
⇔x^2 + 5 x + 4 = 0
⇔x (2 x – 1) + 13 = (x – 3)^2
⇔2 x^2 – x + 13 = (x – 3)^2
⇔2 x^2 – x + 13 = x^2 – 6 x + 9
\(⇔\left[ \begin{array}{l}x=-1\\x=-4\end{array} \right.\)
Vậy S $\in$ {-1;-4}